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Let C be a subset of a compact metric space (X, d). Assume that, for every continuous function h : X → R, the restriction of h to C attains a maximum on C. Prove that C is compact.

My attempt: I intend to show that every infinite subset of C has a limit point in C. Let C' be an infinite subset of C. Since X is compact , C' has a limit point , say q, in X. Suppose q is not in C. Then I am trying to obtain some contradiction to the hypothesis i.e construct a continuous function whose restriction to C has maximum at q. But then q must be in C. I think there is something wrong in the last two lines. But this is what I have tried.

Apart from your own methods, if someone can provide a proof along the lines of the above approach then please do post it.

Thanks.

RagingBull
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  • Is the function $f(x)=1/d(x,q)$ continuous? I'm not sure which is why I am asking. – Eoin Mar 30 '15 at 16:23
  • Is it defined on X or on C? And what is the value at q? – RagingBull Mar 30 '15 at 16:32
  • And I think your function is continuous.See here http://math.stackexchange.com/questions/315306/show-that-a-distance-function-is-continuous and http://math.stackexchange.com/questions/8066/is-the-function-distance-continuous – RagingBull Mar 30 '15 at 16:40
  • Then this function diverges on C but is continuous on C. So every limit point must be in C. Since this implies C is a closed subset of a compact set you have the result. – Eoin Mar 30 '15 at 16:51
  • I get your point. But you have defined the function on C whereas the hypothesis talks about the restriction of any continuous function to C. So in order to be defined on X you just need to define it at q. If I define f(q)=0 will that make f a continuous function on X? – RagingBull Mar 30 '15 at 17:05
  • You're right. But in similar effect could it work to define a function that is continuous on X and obtains a max at $q$? Say $f(x)=1/(1+d(x,q))$. – Eoin Mar 30 '15 at 17:12
  • But how will you show that the restriction to C attains a maximum on C. Thing is d(x,q) can be made arbitrarily small for x in C since q is a limit point, So I think maximum cannot exist on C. What do you think? – RagingBull Mar 30 '15 at 17:23
  • You can go by contradiction after this. By assumption every continuous function on X has this property of restriction. The function f above is a continuous function on X. So it must have this restriction property. But q is a limit point of C' not in C, and this function is a continuous function which violates our assumption. Thus q must be in C. Because f must obtain a max on C. – Eoin Mar 30 '15 at 17:34
  • Ahh nice.Thanks! – RagingBull Mar 30 '15 at 17:36
  • I think f does not violate the hypothesis because q is not in C. f must attain its maximum on C, let that point be p. Then there is no reason that q must be equal to p since we have assumed that q is not in C. Do you get my point? – RagingBull Apr 04 '15 at 15:03

1 Answers1

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Suppose $C$ is a subset of a compact metric space $(X,d)$. Furthermore, assume that every continuous function from $X$ into $\mathbb{R}$ when restricted to $C$ obtains its maximum on $C$.

If $C$ is finite, then $C$ is compact. So we may assume the case $C$ is infinite.

Since $C$ is an infinite subset of the compact space $X$, it has a limit point in $X$, say $q$. Consider the function $f:X\rightarrow \mathbb{R}$ defined by $f(x)=\frac{1}{1+d(x,q)}$. This function obtains its maximum at $d(x,q)=0$ (i.e. $x=q$). Since $q$ is a limit point of $C$, there is a sequence $(x_n)_{n=0}^\infty$ of points in $C$ which converge to $q$. Since $f$, when restricted to $C$, must obtain its maximum by assumption, we have that $q\in C$.

Eoin
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  • I think there are two different situations here: (1) Every continuous function attains its maximum when restricted to C and (2) Every restriction attains its maximum on C. These are two entirely different statements and my original question talks about (2). But your answer talks about (1). I hope you get my point. – RagingBull Apr 04 '15 at 18:59
  • Your assumption is every continuous function defined on $X$, i.e. $f:X\rightarrow \mathbb{R}$, when restricted to $C$, i.e. $f|_C$ obtains its maximum on $C$, or, $f|_C$ has a maximum on $C$ but, since it is restricted to $f|_C:C\rightarrow\mathbb{R}$ it is equivalent to say $f|_C$ obtains its maximum. The function $f$ as defined is a continuous function on $X$. By assumption, $C$ is a subset of $X$ such that $f|_C$ has a maximum. This is only possible if it contains its limit points as stated above. – Eoin Apr 04 '15 at 19:22
  • @BasantSharma The difference is that you are mixing relative maximum to absolute maximum. I do not claim that every continuous function achieves its absolute maximum on $C$. However, the function that I have made obtains an absolute maximum at the limit point, $q$. Since it is a limit point you have that either $q\in C$ or $q\notin C$. In both cases you can construct a sequence which approaches $q$ to within $\epsilon>0$ for all $\epsilon$. As such, this function, $f|_C$ will NOT obtain a maximum, relative or otherwise, if $q\notin C$. This violates your assumption. – Eoin Apr 04 '15 at 19:29
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    If you claim that it obtains a relative maximum on $C$, then you are saying there is some point $x\in C$ with $f(x)=\frac{1}{1+d(x,q)}$ being maximum. If this were true, $q$ could not be a limit point of $C$ since this would mean that there are NO points $y$ with $d(y,q)<d(x,q)$ or I could bound $q$ in some epsilon disk contradicting that it is a limit point. – Eoin Apr 04 '15 at 19:33
  • Now I get it.Thanks a lot :) – RagingBull Apr 05 '15 at 05:21
  • @BasantSharma Awesome! Best of luck. – Eoin Apr 05 '15 at 05:45