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How to prove the triangle inequality for this metric space


Show that in a discrete metric space, every subset is both open and closed.Proof that a discrete metric is indeed a metric spaceContinuity of functions in a metric space.What characteristics of binary space are needed to prove that it is totally disconnected?Continuous functions in a metric space using the discrete metricMetric space - complete - discreteThe limit points of a subset of binary space with ultra metricGiven a distance function $d$ prove it is a metric on the Baire space.How does the metric on a space $X$, affect the way we can define neighborhoods of points $ p in X$?How to prove the set of bounded sequences is clopen in the uniform metric?













2












$begingroup$


Let $X = prod_k=1^infty 0,1$ which can be viewed as the set of all sequences of binary digits. For $f,g in X$, with $f neq g$, define $m(f, g) = mink : f(k) neq g(k)$ and define $d : X times X to mathbbR$ by



$$d(f, g) = begincases2^-m(f,g) & textif f neq g \ 0 & textif f = g.endcases$$



The first three axioms have been given I imagine the first starting point is to prove that



$$m(a,c) ge minm(a,b), m(b,c).$$



Followed by:




For $f in X$, $n in mathbbN$, prove that $B(f, 2^-n)$ consists of all $g in X$ for which
$$f(1) = g(1), f(2) = g(2), ldots, text and f(n) = g(n).$$




This I think you must use the previous fact that $d(f,g) = 2^-m(f,g)$.



Followed by:




Show that for all $f in X$, $r > 0$, either $B'(f,r) = B(f,r)$ or $B'(f,r) = B(f,2r)$.




This I believe must be a continuity using open sets problem.










share|cite|improve this question











$endgroup$







  • 6




    $begingroup$
    Please do mathJax, so difficult to read your question.. math.meta.stackexchange.com/questions/5020/…
    $endgroup$
    – Yujie Zha
    3 hours ago






  • 1




    $begingroup$
    I was going to edit your question to format it properly, but then I noticed that you've been here for almost a year now (happy anniversary, in $5$ days time!) and have asked four prior questions that others have all formatted for you. So, unfortunately, this time I'm just going to down-vote. If you do format your question, give me a ping in the comments (@TheoBendit) and I'll remove the down-vote, attempt to answer the question, and give you an up-vote to boot. If you have any questions about using MathJax, ask Meta. Good luck!
    $endgroup$
    – Theo Bendit
    3 hours ago






  • 1




    $begingroup$
    Alexander, it really does not take much effort, but you have not put it in for 360 days! If you don't know how to do it, then learn it, or keep the MathJax documentation page open while you're typing! I did this for six months , and eventually rote learned all the simple commands. Take the time to do it, because your question ticks all the boxes and deserves attention and upvotes, except for its legibility. (Yes, you are starting to change it, that is good!)
    $endgroup$
    – астон вілла олоф мэллбэрг
    2 hours ago







  • 2




    $begingroup$
    @AlexanderQuinn Removing the down-vote for effort, but I'd like to see a little bit more done before I come in and clean it up for you. You can press "edit" on your previous questions to see how others have formatted them, and you can even try changing things to test how it affects the layout. Here are some simple commands too: neq = "$neq$", in = "$in$", ge = "$ge$". Try putting $ signs around most of the maths, and it will already look much better. If you try another edit, I'll finish it up for you if you wish.
    $endgroup$
    – Theo Bendit
    2 hours ago







  • 2




    $begingroup$
    $What is B'???$
    $endgroup$
    – William Elliot
    2 hours ago















2












$begingroup$


Let $X = prod_k=1^infty 0,1$ which can be viewed as the set of all sequences of binary digits. For $f,g in X$, with $f neq g$, define $m(f, g) = mink : f(k) neq g(k)$ and define $d : X times X to mathbbR$ by



$$d(f, g) = begincases2^-m(f,g) & textif f neq g \ 0 & textif f = g.endcases$$



The first three axioms have been given I imagine the first starting point is to prove that



$$m(a,c) ge minm(a,b), m(b,c).$$



Followed by:




For $f in X$, $n in mathbbN$, prove that $B(f, 2^-n)$ consists of all $g in X$ for which
$$f(1) = g(1), f(2) = g(2), ldots, text and f(n) = g(n).$$




This I think you must use the previous fact that $d(f,g) = 2^-m(f,g)$.



Followed by:




Show that for all $f in X$, $r > 0$, either $B'(f,r) = B(f,r)$ or $B'(f,r) = B(f,2r)$.




This I believe must be a continuity using open sets problem.










share|cite|improve this question











$endgroup$







  • 6




    $begingroup$
    Please do mathJax, so difficult to read your question.. math.meta.stackexchange.com/questions/5020/…
    $endgroup$
    – Yujie Zha
    3 hours ago






  • 1




    $begingroup$
    I was going to edit your question to format it properly, but then I noticed that you've been here for almost a year now (happy anniversary, in $5$ days time!) and have asked four prior questions that others have all formatted for you. So, unfortunately, this time I'm just going to down-vote. If you do format your question, give me a ping in the comments (@TheoBendit) and I'll remove the down-vote, attempt to answer the question, and give you an up-vote to boot. If you have any questions about using MathJax, ask Meta. Good luck!
    $endgroup$
    – Theo Bendit
    3 hours ago






  • 1




    $begingroup$
    Alexander, it really does not take much effort, but you have not put it in for 360 days! If you don't know how to do it, then learn it, or keep the MathJax documentation page open while you're typing! I did this for six months , and eventually rote learned all the simple commands. Take the time to do it, because your question ticks all the boxes and deserves attention and upvotes, except for its legibility. (Yes, you are starting to change it, that is good!)
    $endgroup$
    – астон вілла олоф мэллбэрг
    2 hours ago







  • 2




    $begingroup$
    @AlexanderQuinn Removing the down-vote for effort, but I'd like to see a little bit more done before I come in and clean it up for you. You can press "edit" on your previous questions to see how others have formatted them, and you can even try changing things to test how it affects the layout. Here are some simple commands too: neq = "$neq$", in = "$in$", ge = "$ge$". Try putting $ signs around most of the maths, and it will already look much better. If you try another edit, I'll finish it up for you if you wish.
    $endgroup$
    – Theo Bendit
    2 hours ago







  • 2




    $begingroup$
    $What is B'???$
    $endgroup$
    – William Elliot
    2 hours ago













2












2








2


2



$begingroup$


Let $X = prod_k=1^infty 0,1$ which can be viewed as the set of all sequences of binary digits. For $f,g in X$, with $f neq g$, define $m(f, g) = mink : f(k) neq g(k)$ and define $d : X times X to mathbbR$ by



$$d(f, g) = begincases2^-m(f,g) & textif f neq g \ 0 & textif f = g.endcases$$



The first three axioms have been given I imagine the first starting point is to prove that



$$m(a,c) ge minm(a,b), m(b,c).$$



Followed by:




For $f in X$, $n in mathbbN$, prove that $B(f, 2^-n)$ consists of all $g in X$ for which
$$f(1) = g(1), f(2) = g(2), ldots, text and f(n) = g(n).$$




This I think you must use the previous fact that $d(f,g) = 2^-m(f,g)$.



Followed by:




Show that for all $f in X$, $r > 0$, either $B'(f,r) = B(f,r)$ or $B'(f,r) = B(f,2r)$.




This I believe must be a continuity using open sets problem.










share|cite|improve this question











$endgroup$




Let $X = prod_k=1^infty 0,1$ which can be viewed as the set of all sequences of binary digits. For $f,g in X$, with $f neq g$, define $m(f, g) = mink : f(k) neq g(k)$ and define $d : X times X to mathbbR$ by



$$d(f, g) = begincases2^-m(f,g) & textif f neq g \ 0 & textif f = g.endcases$$



The first three axioms have been given I imagine the first starting point is to prove that



$$m(a,c) ge minm(a,b), m(b,c).$$



Followed by:




For $f in X$, $n in mathbbN$, prove that $B(f, 2^-n)$ consists of all $g in X$ for which
$$f(1) = g(1), f(2) = g(2), ldots, text and f(n) = g(n).$$




This I think you must use the previous fact that $d(f,g) = 2^-m(f,g)$.



Followed by:




Show that for all $f in X$, $r > 0$, either $B'(f,r) = B(f,r)$ or $B'(f,r) = B(f,2r)$.




This I believe must be a continuity using open sets problem.







general-topology metric-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 hours ago









Theo Bendit

19.4k12353




19.4k12353










asked 3 hours ago









Alexander QuinnAlexander Quinn

315




315







  • 6




    $begingroup$
    Please do mathJax, so difficult to read your question.. math.meta.stackexchange.com/questions/5020/…
    $endgroup$
    – Yujie Zha
    3 hours ago






  • 1




    $begingroup$
    I was going to edit your question to format it properly, but then I noticed that you've been here for almost a year now (happy anniversary, in $5$ days time!) and have asked four prior questions that others have all formatted for you. So, unfortunately, this time I'm just going to down-vote. If you do format your question, give me a ping in the comments (@TheoBendit) and I'll remove the down-vote, attempt to answer the question, and give you an up-vote to boot. If you have any questions about using MathJax, ask Meta. Good luck!
    $endgroup$
    – Theo Bendit
    3 hours ago






  • 1




    $begingroup$
    Alexander, it really does not take much effort, but you have not put it in for 360 days! If you don't know how to do it, then learn it, or keep the MathJax documentation page open while you're typing! I did this for six months , and eventually rote learned all the simple commands. Take the time to do it, because your question ticks all the boxes and deserves attention and upvotes, except for its legibility. (Yes, you are starting to change it, that is good!)
    $endgroup$
    – астон вілла олоф мэллбэрг
    2 hours ago







  • 2




    $begingroup$
    @AlexanderQuinn Removing the down-vote for effort, but I'd like to see a little bit more done before I come in and clean it up for you. You can press "edit" on your previous questions to see how others have formatted them, and you can even try changing things to test how it affects the layout. Here are some simple commands too: neq = "$neq$", in = "$in$", ge = "$ge$". Try putting $ signs around most of the maths, and it will already look much better. If you try another edit, I'll finish it up for you if you wish.
    $endgroup$
    – Theo Bendit
    2 hours ago







  • 2




    $begingroup$
    $What is B'???$
    $endgroup$
    – William Elliot
    2 hours ago












  • 6




    $begingroup$
    Please do mathJax, so difficult to read your question.. math.meta.stackexchange.com/questions/5020/…
    $endgroup$
    – Yujie Zha
    3 hours ago






  • 1




    $begingroup$
    I was going to edit your question to format it properly, but then I noticed that you've been here for almost a year now (happy anniversary, in $5$ days time!) and have asked four prior questions that others have all formatted for you. So, unfortunately, this time I'm just going to down-vote. If you do format your question, give me a ping in the comments (@TheoBendit) and I'll remove the down-vote, attempt to answer the question, and give you an up-vote to boot. If you have any questions about using MathJax, ask Meta. Good luck!
    $endgroup$
    – Theo Bendit
    3 hours ago






  • 1




    $begingroup$
    Alexander, it really does not take much effort, but you have not put it in for 360 days! If you don't know how to do it, then learn it, or keep the MathJax documentation page open while you're typing! I did this for six months , and eventually rote learned all the simple commands. Take the time to do it, because your question ticks all the boxes and deserves attention and upvotes, except for its legibility. (Yes, you are starting to change it, that is good!)
    $endgroup$
    – астон вілла олоф мэллбэрг
    2 hours ago







  • 2




    $begingroup$
    @AlexanderQuinn Removing the down-vote for effort, but I'd like to see a little bit more done before I come in and clean it up for you. You can press "edit" on your previous questions to see how others have formatted them, and you can even try changing things to test how it affects the layout. Here are some simple commands too: neq = "$neq$", in = "$in$", ge = "$ge$". Try putting $ signs around most of the maths, and it will already look much better. If you try another edit, I'll finish it up for you if you wish.
    $endgroup$
    – Theo Bendit
    2 hours ago







  • 2




    $begingroup$
    $What is B'???$
    $endgroup$
    – William Elliot
    2 hours ago







6




6




$begingroup$
Please do mathJax, so difficult to read your question.. math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Yujie Zha
3 hours ago




$begingroup$
Please do mathJax, so difficult to read your question.. math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Yujie Zha
3 hours ago




1




1




$begingroup$
I was going to edit your question to format it properly, but then I noticed that you've been here for almost a year now (happy anniversary, in $5$ days time!) and have asked four prior questions that others have all formatted for you. So, unfortunately, this time I'm just going to down-vote. If you do format your question, give me a ping in the comments (@TheoBendit) and I'll remove the down-vote, attempt to answer the question, and give you an up-vote to boot. If you have any questions about using MathJax, ask Meta. Good luck!
$endgroup$
– Theo Bendit
3 hours ago




$begingroup$
I was going to edit your question to format it properly, but then I noticed that you've been here for almost a year now (happy anniversary, in $5$ days time!) and have asked four prior questions that others have all formatted for you. So, unfortunately, this time I'm just going to down-vote. If you do format your question, give me a ping in the comments (@TheoBendit) and I'll remove the down-vote, attempt to answer the question, and give you an up-vote to boot. If you have any questions about using MathJax, ask Meta. Good luck!
$endgroup$
– Theo Bendit
3 hours ago




1




1




$begingroup$
Alexander, it really does not take much effort, but you have not put it in for 360 days! If you don't know how to do it, then learn it, or keep the MathJax documentation page open while you're typing! I did this for six months , and eventually rote learned all the simple commands. Take the time to do it, because your question ticks all the boxes and deserves attention and upvotes, except for its legibility. (Yes, you are starting to change it, that is good!)
$endgroup$
– астон вілла олоф мэллбэрг
2 hours ago





$begingroup$
Alexander, it really does not take much effort, but you have not put it in for 360 days! If you don't know how to do it, then learn it, or keep the MathJax documentation page open while you're typing! I did this for six months , and eventually rote learned all the simple commands. Take the time to do it, because your question ticks all the boxes and deserves attention and upvotes, except for its legibility. (Yes, you are starting to change it, that is good!)
$endgroup$
– астон вілла олоф мэллбэрг
2 hours ago





2




2




$begingroup$
@AlexanderQuinn Removing the down-vote for effort, but I'd like to see a little bit more done before I come in and clean it up for you. You can press "edit" on your previous questions to see how others have formatted them, and you can even try changing things to test how it affects the layout. Here are some simple commands too: neq = "$neq$", in = "$in$", ge = "$ge$". Try putting $ signs around most of the maths, and it will already look much better. If you try another edit, I'll finish it up for you if you wish.
$endgroup$
– Theo Bendit
2 hours ago





$begingroup$
@AlexanderQuinn Removing the down-vote for effort, but I'd like to see a little bit more done before I come in and clean it up for you. You can press "edit" on your previous questions to see how others have formatted them, and you can even try changing things to test how it affects the layout. Here are some simple commands too: neq = "$neq$", in = "$in$", ge = "$ge$". Try putting $ signs around most of the maths, and it will already look much better. If you try another edit, I'll finish it up for you if you wish.
$endgroup$
– Theo Bendit
2 hours ago





2




2




$begingroup$
$What is B'???$
$endgroup$
– William Elliot
2 hours ago




$begingroup$
$What is B'???$
$endgroup$
– William Elliot
2 hours ago










3 Answers
3






active

oldest

votes


















3












$begingroup$

Suppose $f ne g ne h ne f.$ Let $a=m(f,g),,b=m(g,h), ,c=m(f,h).$



Suppose $d(f,g)+d(g,h)<d(f,h).$



Then $ 2^-a+2^-b<2^-c,$ which implies $(,c<a land c<b,). $ But then $$ f(c)=g(c)quad (text as c<a=m(f,g),)$$ $$ text and quad g(c)=h(c)quad (text as c<b=m(g,h),)$$ so $f(c)=h(c),$ contrary to the def'n of $c=m(f,h).$ Contradiction.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    Thanks! So then perhaps the second part can be done by contradiction also?
    $endgroup$
    – Alexander Quinn
    1 hour ago










  • $begingroup$
    This is a very nice answer.
    $endgroup$
    – H. H.
    45 mins ago


















2












$begingroup$

If $a(i)$ agrees with $b(i)$ for all $i = 1, ldots, m(a, b) - 1$. Similarly, $b(i)$ agrees with $c(i)$ for all $i = 1, ldots, m(b, c) - 1$. So, we have $a(i) = b(i) = c(i)$ for all $i = 1, ldots, k$, where $k = min m(a, b), m(b, c) - 1$, as $k$ is a number smaller than both $m(a, b) - 1$ and $m(b, c) - 1$. Therefore, $m(a, c)$, the first $i$ such that $a(i) neq c(i)$, must be at least $k + 1$. That is,
$$m(a, c) ge minm(a, b), m(b, c).$$



Thus, excluding the easy cases where $a, b, c$ are not pairwise distinct,
beginalign*
d(a, b) + d(b, c) &= 2^-m(a, b) + 2^-m(b, c) \
&= 2^-maxm(a, b), m(b, c) + 2^-minm(a, b), (b, c) \
&ge 2^-minm(a, b), (b, c) \
&ge 2^-m(a, c) = d(a, c).
endalign*




If $g in B(f, 2^-n)$, then either $f = g$ (in which case, the result is trivially true), or
$$2^-n > d(f, g) = 2^-m(a, b) implies m(a, b) > n.$$
So, the first point at which $f$ and $g$ disagree occurs past $n$, i.e. $f(i) = g(i)$ for $i = 1, ldots, n$.




Note that the distance function $d(f, g)$ only takes countably many values:
$$0, frac12, frac14, ldots, frac12^n, ldots$$
When $r$ is between these values, the ball $B(f; r)$ will not grow or shrink when you vary $r$ by a small amount. For example, for any $f in X$,
$$Bleft(f, frac13right) = Bleft(f, frac12right) = B'left(f, frac14right),$$
simply because a point that at most $frac13$ distance away from $f$, basically only rules out their distances being $frac12$.



This might give you an idea about how to go about attacking this third problem. I'll let you think about it. Let me know via comment if you need further help.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    So the main thing in the last question is that we need to prove for any f ∈ X and radius r, the closure of the open ball of radius r around x is the closed ball of radius r.
    $endgroup$
    – Alexander Quinn
    1 hour ago










  • $begingroup$
    @AlexanderQuinn Well actually, every open ball is closed, and is equal to a closed ball! But. not necessarily the closed ball of the same radius as the open ball. The closure of $B(f, 1/2)$ is $B(f, 1/2) = B'(f, 1/4)$, not $B(f, 1/2)$ (which is equal to $X$).
    $endgroup$
    – Theo Bendit
    52 mins ago











  • $begingroup$
    So it must be the B(f,r) = B'(f,r) is the only reasonable one out of the two as if you take the limit you eventually approach 0 for r
    $endgroup$
    – Alexander Quinn
    42 mins ago










  • $begingroup$
    (In my previous comment, the third (and last) instance of $B(f, 1/2)$ should be $B'(f, 1/2)$)
    $endgroup$
    – Theo Bendit
    40 mins ago










  • $begingroup$
    @AlexanderQuinn I'm not certain what you mean here. You will have $B(f, r) = B'(f, r)$ when $r$ is not $0$ or a negative power of $2$, as in this case, the difference between the sets (which is the sphere $S(f, r)$) is empty! If $r$ is not one of these numbers, then there are no points in the space of distance exactly $r$ from each other.
    $endgroup$
    – Theo Bendit
    38 mins ago


















0












$begingroup$

For any $ain X$ and radius $r$, the closure of the open ball of radius $r$ around $a$ is the closed ball of radius $r$.



For any two distinct points $a,b$ in the space and any positive $epsilon$, there is a point $c$ within $epsilon$ of $b$, and closer to $a$ than $b$ is.
That is, for every $aneq b$ and $epsilongt 0$, there is $c$ with $d(c,b)<epsilon$ and $d(a,c)<d(a,b)$.



Proof. If the closed ball property holds, then fix any $a,b$ with $r=d(a,b)$. Since the closure of $B$ includes $b$, the second property follows. Conversely, if the second property holds, then if $r=d(a,b)$, then the property ensures that $b$ is in the closure of $B$, and so the closure of the open ball includes the closed ball






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    3 Answers
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    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

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    active

    oldest

    votes









    3












    $begingroup$

    Suppose $f ne g ne h ne f.$ Let $a=m(f,g),,b=m(g,h), ,c=m(f,h).$



    Suppose $d(f,g)+d(g,h)<d(f,h).$



    Then $ 2^-a+2^-b<2^-c,$ which implies $(,c<a land c<b,). $ But then $$ f(c)=g(c)quad (text as c<a=m(f,g),)$$ $$ text and quad g(c)=h(c)quad (text as c<b=m(g,h),)$$ so $f(c)=h(c),$ contrary to the def'n of $c=m(f,h).$ Contradiction.






    share|cite|improve this answer











    $endgroup$












    • $begingroup$
      Thanks! So then perhaps the second part can be done by contradiction also?
      $endgroup$
      – Alexander Quinn
      1 hour ago










    • $begingroup$
      This is a very nice answer.
      $endgroup$
      – H. H.
      45 mins ago















    3












    $begingroup$

    Suppose $f ne g ne h ne f.$ Let $a=m(f,g),,b=m(g,h), ,c=m(f,h).$



    Suppose $d(f,g)+d(g,h)<d(f,h).$



    Then $ 2^-a+2^-b<2^-c,$ which implies $(,c<a land c<b,). $ But then $$ f(c)=g(c)quad (text as c<a=m(f,g),)$$ $$ text and quad g(c)=h(c)quad (text as c<b=m(g,h),)$$ so $f(c)=h(c),$ contrary to the def'n of $c=m(f,h).$ Contradiction.






    share|cite|improve this answer











    $endgroup$












    • $begingroup$
      Thanks! So then perhaps the second part can be done by contradiction also?
      $endgroup$
      – Alexander Quinn
      1 hour ago










    • $begingroup$
      This is a very nice answer.
      $endgroup$
      – H. H.
      45 mins ago













    3












    3








    3





    $begingroup$

    Suppose $f ne g ne h ne f.$ Let $a=m(f,g),,b=m(g,h), ,c=m(f,h).$



    Suppose $d(f,g)+d(g,h)<d(f,h).$



    Then $ 2^-a+2^-b<2^-c,$ which implies $(,c<a land c<b,). $ But then $$ f(c)=g(c)quad (text as c<a=m(f,g),)$$ $$ text and quad g(c)=h(c)quad (text as c<b=m(g,h),)$$ so $f(c)=h(c),$ contrary to the def'n of $c=m(f,h).$ Contradiction.






    share|cite|improve this answer











    $endgroup$



    Suppose $f ne g ne h ne f.$ Let $a=m(f,g),,b=m(g,h), ,c=m(f,h).$



    Suppose $d(f,g)+d(g,h)<d(f,h).$



    Then $ 2^-a+2^-b<2^-c,$ which implies $(,c<a land c<b,). $ But then $$ f(c)=g(c)quad (text as c<a=m(f,g),)$$ $$ text and quad g(c)=h(c)quad (text as c<b=m(g,h),)$$ so $f(c)=h(c),$ contrary to the def'n of $c=m(f,h).$ Contradiction.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited 1 hour ago

























    answered 1 hour ago









    DanielWainfleetDanielWainfleet

    35.5k31648




    35.5k31648











    • $begingroup$
      Thanks! So then perhaps the second part can be done by contradiction also?
      $endgroup$
      – Alexander Quinn
      1 hour ago










    • $begingroup$
      This is a very nice answer.
      $endgroup$
      – H. H.
      45 mins ago
















    • $begingroup$
      Thanks! So then perhaps the second part can be done by contradiction also?
      $endgroup$
      – Alexander Quinn
      1 hour ago










    • $begingroup$
      This is a very nice answer.
      $endgroup$
      – H. H.
      45 mins ago















    $begingroup$
    Thanks! So then perhaps the second part can be done by contradiction also?
    $endgroup$
    – Alexander Quinn
    1 hour ago




    $begingroup$
    Thanks! So then perhaps the second part can be done by contradiction also?
    $endgroup$
    – Alexander Quinn
    1 hour ago












    $begingroup$
    This is a very nice answer.
    $endgroup$
    – H. H.
    45 mins ago




    $begingroup$
    This is a very nice answer.
    $endgroup$
    – H. H.
    45 mins ago











    2












    $begingroup$

    If $a(i)$ agrees with $b(i)$ for all $i = 1, ldots, m(a, b) - 1$. Similarly, $b(i)$ agrees with $c(i)$ for all $i = 1, ldots, m(b, c) - 1$. So, we have $a(i) = b(i) = c(i)$ for all $i = 1, ldots, k$, where $k = min m(a, b), m(b, c) - 1$, as $k$ is a number smaller than both $m(a, b) - 1$ and $m(b, c) - 1$. Therefore, $m(a, c)$, the first $i$ such that $a(i) neq c(i)$, must be at least $k + 1$. That is,
    $$m(a, c) ge minm(a, b), m(b, c).$$



    Thus, excluding the easy cases where $a, b, c$ are not pairwise distinct,
    beginalign*
    d(a, b) + d(b, c) &= 2^-m(a, b) + 2^-m(b, c) \
    &= 2^-maxm(a, b), m(b, c) + 2^-minm(a, b), (b, c) \
    &ge 2^-minm(a, b), (b, c) \
    &ge 2^-m(a, c) = d(a, c).
    endalign*




    If $g in B(f, 2^-n)$, then either $f = g$ (in which case, the result is trivially true), or
    $$2^-n > d(f, g) = 2^-m(a, b) implies m(a, b) > n.$$
    So, the first point at which $f$ and $g$ disagree occurs past $n$, i.e. $f(i) = g(i)$ for $i = 1, ldots, n$.




    Note that the distance function $d(f, g)$ only takes countably many values:
    $$0, frac12, frac14, ldots, frac12^n, ldots$$
    When $r$ is between these values, the ball $B(f; r)$ will not grow or shrink when you vary $r$ by a small amount. For example, for any $f in X$,
    $$Bleft(f, frac13right) = Bleft(f, frac12right) = B'left(f, frac14right),$$
    simply because a point that at most $frac13$ distance away from $f$, basically only rules out their distances being $frac12$.



    This might give you an idea about how to go about attacking this third problem. I'll let you think about it. Let me know via comment if you need further help.






    share|cite|improve this answer









    $endgroup$












    • $begingroup$
      So the main thing in the last question is that we need to prove for any f ∈ X and radius r, the closure of the open ball of radius r around x is the closed ball of radius r.
      $endgroup$
      – Alexander Quinn
      1 hour ago










    • $begingroup$
      @AlexanderQuinn Well actually, every open ball is closed, and is equal to a closed ball! But. not necessarily the closed ball of the same radius as the open ball. The closure of $B(f, 1/2)$ is $B(f, 1/2) = B'(f, 1/4)$, not $B(f, 1/2)$ (which is equal to $X$).
      $endgroup$
      – Theo Bendit
      52 mins ago











    • $begingroup$
      So it must be the B(f,r) = B'(f,r) is the only reasonable one out of the two as if you take the limit you eventually approach 0 for r
      $endgroup$
      – Alexander Quinn
      42 mins ago










    • $begingroup$
      (In my previous comment, the third (and last) instance of $B(f, 1/2)$ should be $B'(f, 1/2)$)
      $endgroup$
      – Theo Bendit
      40 mins ago










    • $begingroup$
      @AlexanderQuinn I'm not certain what you mean here. You will have $B(f, r) = B'(f, r)$ when $r$ is not $0$ or a negative power of $2$, as in this case, the difference between the sets (which is the sphere $S(f, r)$) is empty! If $r$ is not one of these numbers, then there are no points in the space of distance exactly $r$ from each other.
      $endgroup$
      – Theo Bendit
      38 mins ago















    2












    $begingroup$

    If $a(i)$ agrees with $b(i)$ for all $i = 1, ldots, m(a, b) - 1$. Similarly, $b(i)$ agrees with $c(i)$ for all $i = 1, ldots, m(b, c) - 1$. So, we have $a(i) = b(i) = c(i)$ for all $i = 1, ldots, k$, where $k = min m(a, b), m(b, c) - 1$, as $k$ is a number smaller than both $m(a, b) - 1$ and $m(b, c) - 1$. Therefore, $m(a, c)$, the first $i$ such that $a(i) neq c(i)$, must be at least $k + 1$. That is,
    $$m(a, c) ge minm(a, b), m(b, c).$$



    Thus, excluding the easy cases where $a, b, c$ are not pairwise distinct,
    beginalign*
    d(a, b) + d(b, c) &= 2^-m(a, b) + 2^-m(b, c) \
    &= 2^-maxm(a, b), m(b, c) + 2^-minm(a, b), (b, c) \
    &ge 2^-minm(a, b), (b, c) \
    &ge 2^-m(a, c) = d(a, c).
    endalign*




    If $g in B(f, 2^-n)$, then either $f = g$ (in which case, the result is trivially true), or
    $$2^-n > d(f, g) = 2^-m(a, b) implies m(a, b) > n.$$
    So, the first point at which $f$ and $g$ disagree occurs past $n$, i.e. $f(i) = g(i)$ for $i = 1, ldots, n$.




    Note that the distance function $d(f, g)$ only takes countably many values:
    $$0, frac12, frac14, ldots, frac12^n, ldots$$
    When $r$ is between these values, the ball $B(f; r)$ will not grow or shrink when you vary $r$ by a small amount. For example, for any $f in X$,
    $$Bleft(f, frac13right) = Bleft(f, frac12right) = B'left(f, frac14right),$$
    simply because a point that at most $frac13$ distance away from $f$, basically only rules out their distances being $frac12$.



    This might give you an idea about how to go about attacking this third problem. I'll let you think about it. Let me know via comment if you need further help.






    share|cite|improve this answer









    $endgroup$












    • $begingroup$
      So the main thing in the last question is that we need to prove for any f ∈ X and radius r, the closure of the open ball of radius r around x is the closed ball of radius r.
      $endgroup$
      – Alexander Quinn
      1 hour ago










    • $begingroup$
      @AlexanderQuinn Well actually, every open ball is closed, and is equal to a closed ball! But. not necessarily the closed ball of the same radius as the open ball. The closure of $B(f, 1/2)$ is $B(f, 1/2) = B'(f, 1/4)$, not $B(f, 1/2)$ (which is equal to $X$).
      $endgroup$
      – Theo Bendit
      52 mins ago











    • $begingroup$
      So it must be the B(f,r) = B'(f,r) is the only reasonable one out of the two as if you take the limit you eventually approach 0 for r
      $endgroup$
      – Alexander Quinn
      42 mins ago










    • $begingroup$
      (In my previous comment, the third (and last) instance of $B(f, 1/2)$ should be $B'(f, 1/2)$)
      $endgroup$
      – Theo Bendit
      40 mins ago










    • $begingroup$
      @AlexanderQuinn I'm not certain what you mean here. You will have $B(f, r) = B'(f, r)$ when $r$ is not $0$ or a negative power of $2$, as in this case, the difference between the sets (which is the sphere $S(f, r)$) is empty! If $r$ is not one of these numbers, then there are no points in the space of distance exactly $r$ from each other.
      $endgroup$
      – Theo Bendit
      38 mins ago













    2












    2








    2





    $begingroup$

    If $a(i)$ agrees with $b(i)$ for all $i = 1, ldots, m(a, b) - 1$. Similarly, $b(i)$ agrees with $c(i)$ for all $i = 1, ldots, m(b, c) - 1$. So, we have $a(i) = b(i) = c(i)$ for all $i = 1, ldots, k$, where $k = min m(a, b), m(b, c) - 1$, as $k$ is a number smaller than both $m(a, b) - 1$ and $m(b, c) - 1$. Therefore, $m(a, c)$, the first $i$ such that $a(i) neq c(i)$, must be at least $k + 1$. That is,
    $$m(a, c) ge minm(a, b), m(b, c).$$



    Thus, excluding the easy cases where $a, b, c$ are not pairwise distinct,
    beginalign*
    d(a, b) + d(b, c) &= 2^-m(a, b) + 2^-m(b, c) \
    &= 2^-maxm(a, b), m(b, c) + 2^-minm(a, b), (b, c) \
    &ge 2^-minm(a, b), (b, c) \
    &ge 2^-m(a, c) = d(a, c).
    endalign*




    If $g in B(f, 2^-n)$, then either $f = g$ (in which case, the result is trivially true), or
    $$2^-n > d(f, g) = 2^-m(a, b) implies m(a, b) > n.$$
    So, the first point at which $f$ and $g$ disagree occurs past $n$, i.e. $f(i) = g(i)$ for $i = 1, ldots, n$.




    Note that the distance function $d(f, g)$ only takes countably many values:
    $$0, frac12, frac14, ldots, frac12^n, ldots$$
    When $r$ is between these values, the ball $B(f; r)$ will not grow or shrink when you vary $r$ by a small amount. For example, for any $f in X$,
    $$Bleft(f, frac13right) = Bleft(f, frac12right) = B'left(f, frac14right),$$
    simply because a point that at most $frac13$ distance away from $f$, basically only rules out their distances being $frac12$.



    This might give you an idea about how to go about attacking this third problem. I'll let you think about it. Let me know via comment if you need further help.






    share|cite|improve this answer









    $endgroup$



    If $a(i)$ agrees with $b(i)$ for all $i = 1, ldots, m(a, b) - 1$. Similarly, $b(i)$ agrees with $c(i)$ for all $i = 1, ldots, m(b, c) - 1$. So, we have $a(i) = b(i) = c(i)$ for all $i = 1, ldots, k$, where $k = min m(a, b), m(b, c) - 1$, as $k$ is a number smaller than both $m(a, b) - 1$ and $m(b, c) - 1$. Therefore, $m(a, c)$, the first $i$ such that $a(i) neq c(i)$, must be at least $k + 1$. That is,
    $$m(a, c) ge minm(a, b), m(b, c).$$



    Thus, excluding the easy cases where $a, b, c$ are not pairwise distinct,
    beginalign*
    d(a, b) + d(b, c) &= 2^-m(a, b) + 2^-m(b, c) \
    &= 2^-maxm(a, b), m(b, c) + 2^-minm(a, b), (b, c) \
    &ge 2^-minm(a, b), (b, c) \
    &ge 2^-m(a, c) = d(a, c).
    endalign*




    If $g in B(f, 2^-n)$, then either $f = g$ (in which case, the result is trivially true), or
    $$2^-n > d(f, g) = 2^-m(a, b) implies m(a, b) > n.$$
    So, the first point at which $f$ and $g$ disagree occurs past $n$, i.e. $f(i) = g(i)$ for $i = 1, ldots, n$.




    Note that the distance function $d(f, g)$ only takes countably many values:
    $$0, frac12, frac14, ldots, frac12^n, ldots$$
    When $r$ is between these values, the ball $B(f; r)$ will not grow or shrink when you vary $r$ by a small amount. For example, for any $f in X$,
    $$Bleft(f, frac13right) = Bleft(f, frac12right) = B'left(f, frac14right),$$
    simply because a point that at most $frac13$ distance away from $f$, basically only rules out their distances being $frac12$.



    This might give you an idea about how to go about attacking this third problem. I'll let you think about it. Let me know via comment if you need further help.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 1 hour ago









    Theo BenditTheo Bendit

    19.4k12353




    19.4k12353











    • $begingroup$
      So the main thing in the last question is that we need to prove for any f ∈ X and radius r, the closure of the open ball of radius r around x is the closed ball of radius r.
      $endgroup$
      – Alexander Quinn
      1 hour ago










    • $begingroup$
      @AlexanderQuinn Well actually, every open ball is closed, and is equal to a closed ball! But. not necessarily the closed ball of the same radius as the open ball. The closure of $B(f, 1/2)$ is $B(f, 1/2) = B'(f, 1/4)$, not $B(f, 1/2)$ (which is equal to $X$).
      $endgroup$
      – Theo Bendit
      52 mins ago











    • $begingroup$
      So it must be the B(f,r) = B'(f,r) is the only reasonable one out of the two as if you take the limit you eventually approach 0 for r
      $endgroup$
      – Alexander Quinn
      42 mins ago










    • $begingroup$
      (In my previous comment, the third (and last) instance of $B(f, 1/2)$ should be $B'(f, 1/2)$)
      $endgroup$
      – Theo Bendit
      40 mins ago










    • $begingroup$
      @AlexanderQuinn I'm not certain what you mean here. You will have $B(f, r) = B'(f, r)$ when $r$ is not $0$ or a negative power of $2$, as in this case, the difference between the sets (which is the sphere $S(f, r)$) is empty! If $r$ is not one of these numbers, then there are no points in the space of distance exactly $r$ from each other.
      $endgroup$
      – Theo Bendit
      38 mins ago
















    • $begingroup$
      So the main thing in the last question is that we need to prove for any f ∈ X and radius r, the closure of the open ball of radius r around x is the closed ball of radius r.
      $endgroup$
      – Alexander Quinn
      1 hour ago










    • $begingroup$
      @AlexanderQuinn Well actually, every open ball is closed, and is equal to a closed ball! But. not necessarily the closed ball of the same radius as the open ball. The closure of $B(f, 1/2)$ is $B(f, 1/2) = B'(f, 1/4)$, not $B(f, 1/2)$ (which is equal to $X$).
      $endgroup$
      – Theo Bendit
      52 mins ago











    • $begingroup$
      So it must be the B(f,r) = B'(f,r) is the only reasonable one out of the two as if you take the limit you eventually approach 0 for r
      $endgroup$
      – Alexander Quinn
      42 mins ago










    • $begingroup$
      (In my previous comment, the third (and last) instance of $B(f, 1/2)$ should be $B'(f, 1/2)$)
      $endgroup$
      – Theo Bendit
      40 mins ago










    • $begingroup$
      @AlexanderQuinn I'm not certain what you mean here. You will have $B(f, r) = B'(f, r)$ when $r$ is not $0$ or a negative power of $2$, as in this case, the difference between the sets (which is the sphere $S(f, r)$) is empty! If $r$ is not one of these numbers, then there are no points in the space of distance exactly $r$ from each other.
      $endgroup$
      – Theo Bendit
      38 mins ago















    $begingroup$
    So the main thing in the last question is that we need to prove for any f ∈ X and radius r, the closure of the open ball of radius r around x is the closed ball of radius r.
    $endgroup$
    – Alexander Quinn
    1 hour ago




    $begingroup$
    So the main thing in the last question is that we need to prove for any f ∈ X and radius r, the closure of the open ball of radius r around x is the closed ball of radius r.
    $endgroup$
    – Alexander Quinn
    1 hour ago












    $begingroup$
    @AlexanderQuinn Well actually, every open ball is closed, and is equal to a closed ball! But. not necessarily the closed ball of the same radius as the open ball. The closure of $B(f, 1/2)$ is $B(f, 1/2) = B'(f, 1/4)$, not $B(f, 1/2)$ (which is equal to $X$).
    $endgroup$
    – Theo Bendit
    52 mins ago





    $begingroup$
    @AlexanderQuinn Well actually, every open ball is closed, and is equal to a closed ball! But. not necessarily the closed ball of the same radius as the open ball. The closure of $B(f, 1/2)$ is $B(f, 1/2) = B'(f, 1/4)$, not $B(f, 1/2)$ (which is equal to $X$).
    $endgroup$
    – Theo Bendit
    52 mins ago













    $begingroup$
    So it must be the B(f,r) = B'(f,r) is the only reasonable one out of the two as if you take the limit you eventually approach 0 for r
    $endgroup$
    – Alexander Quinn
    42 mins ago




    $begingroup$
    So it must be the B(f,r) = B'(f,r) is the only reasonable one out of the two as if you take the limit you eventually approach 0 for r
    $endgroup$
    – Alexander Quinn
    42 mins ago












    $begingroup$
    (In my previous comment, the third (and last) instance of $B(f, 1/2)$ should be $B'(f, 1/2)$)
    $endgroup$
    – Theo Bendit
    40 mins ago




    $begingroup$
    (In my previous comment, the third (and last) instance of $B(f, 1/2)$ should be $B'(f, 1/2)$)
    $endgroup$
    – Theo Bendit
    40 mins ago












    $begingroup$
    @AlexanderQuinn I'm not certain what you mean here. You will have $B(f, r) = B'(f, r)$ when $r$ is not $0$ or a negative power of $2$, as in this case, the difference between the sets (which is the sphere $S(f, r)$) is empty! If $r$ is not one of these numbers, then there are no points in the space of distance exactly $r$ from each other.
    $endgroup$
    – Theo Bendit
    38 mins ago




    $begingroup$
    @AlexanderQuinn I'm not certain what you mean here. You will have $B(f, r) = B'(f, r)$ when $r$ is not $0$ or a negative power of $2$, as in this case, the difference between the sets (which is the sphere $S(f, r)$) is empty! If $r$ is not one of these numbers, then there are no points in the space of distance exactly $r$ from each other.
    $endgroup$
    – Theo Bendit
    38 mins ago











    0












    $begingroup$

    For any $ain X$ and radius $r$, the closure of the open ball of radius $r$ around $a$ is the closed ball of radius $r$.



    For any two distinct points $a,b$ in the space and any positive $epsilon$, there is a point $c$ within $epsilon$ of $b$, and closer to $a$ than $b$ is.
    That is, for every $aneq b$ and $epsilongt 0$, there is $c$ with $d(c,b)<epsilon$ and $d(a,c)<d(a,b)$.



    Proof. If the closed ball property holds, then fix any $a,b$ with $r=d(a,b)$. Since the closure of $B$ includes $b$, the second property follows. Conversely, if the second property holds, then if $r=d(a,b)$, then the property ensures that $b$ is in the closure of $B$, and so the closure of the open ball includes the closed ball






    share|cite|improve this answer









    $endgroup$

















      0












      $begingroup$

      For any $ain X$ and radius $r$, the closure of the open ball of radius $r$ around $a$ is the closed ball of radius $r$.



      For any two distinct points $a,b$ in the space and any positive $epsilon$, there is a point $c$ within $epsilon$ of $b$, and closer to $a$ than $b$ is.
      That is, for every $aneq b$ and $epsilongt 0$, there is $c$ with $d(c,b)<epsilon$ and $d(a,c)<d(a,b)$.



      Proof. If the closed ball property holds, then fix any $a,b$ with $r=d(a,b)$. Since the closure of $B$ includes $b$, the second property follows. Conversely, if the second property holds, then if $r=d(a,b)$, then the property ensures that $b$ is in the closure of $B$, and so the closure of the open ball includes the closed ball






      share|cite|improve this answer









      $endgroup$















        0












        0








        0





        $begingroup$

        For any $ain X$ and radius $r$, the closure of the open ball of radius $r$ around $a$ is the closed ball of radius $r$.



        For any two distinct points $a,b$ in the space and any positive $epsilon$, there is a point $c$ within $epsilon$ of $b$, and closer to $a$ than $b$ is.
        That is, for every $aneq b$ and $epsilongt 0$, there is $c$ with $d(c,b)<epsilon$ and $d(a,c)<d(a,b)$.



        Proof. If the closed ball property holds, then fix any $a,b$ with $r=d(a,b)$. Since the closure of $B$ includes $b$, the second property follows. Conversely, if the second property holds, then if $r=d(a,b)$, then the property ensures that $b$ is in the closure of $B$, and so the closure of the open ball includes the closed ball






        share|cite|improve this answer









        $endgroup$



        For any $ain X$ and radius $r$, the closure of the open ball of radius $r$ around $a$ is the closed ball of radius $r$.



        For any two distinct points $a,b$ in the space and any positive $epsilon$, there is a point $c$ within $epsilon$ of $b$, and closer to $a$ than $b$ is.
        That is, for every $aneq b$ and $epsilongt 0$, there is $c$ with $d(c,b)<epsilon$ and $d(a,c)<d(a,b)$.



        Proof. If the closed ball property holds, then fix any $a,b$ with $r=d(a,b)$. Since the closure of $B$ includes $b$, the second property follows. Conversely, if the second property holds, then if $r=d(a,b)$, then the property ensures that $b$ is in the closure of $B$, and so the closure of the open ball includes the closed ball







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        answered 59 mins ago









        Alexander QuinnAlexander Quinn

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