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To prove:

"If every continuous function $f$ in $X \subset \mathbb{R}^2$ is bounded then $X$ is compact."

My attempt :

In $\mathbb{R}^n$ a set $X$ is compact iff it is closed and bounded. I can show $X$ is bounded, but can not prove $X$ is closed.

The distance function $f : X \rightarrow \mathbb{R}$ defined by $f(x , y) = \sqrt{x^2 + y^2}$ is continuous on $X$. So it is bounded. Thus distance between any two points of $X$ is finite and hence $X$ is bounded.

Let $a \in \mathbb{R}^2$ be a limit point of $X$. Consider a sequence $\{a_n\}$ in $X$ converges to $a$. So for any continuous function $f$, the sequence $\{f(a_n)\}$ converges to $f(a)$ and $f(a)$ is finite. It does not imply $a \in X$.

Thank you for your help.

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Mr.Prahlad's answer for my question… might be helpful for you I believe. – Praphulla Koushik Nov 23 '13 at 6:17
@Samprity : To show that $X$ is bounded, by definition you must show that $X \subseteq \left\{ (x,y) \in \mathbb R^2 \, \left| \, \sqrt{(x-x_0)^2 + (y-y_0)^2} < R \right. \right\}$ for a certain $R > 0$ and $(x_0,y_0)$ (this set is often abbreviated $B_R(x_0,y_0)$. Your argument precisely shows that $X$ is a subset of such a set for $x_0=y_0=0$ and $R$ large enough. Logically speaking, "Thus distance between any two points of $X$ is finite" is a consequence of the boundedness of $X$, not the other way around. :) – Patrick Da Silva Nov 23 '13 at 6:33
@PraphullaKoushik Yes Mr. Vidyanathan's answer is helpful. – Dutta Nov 23 '13 at 6:55
up vote 3 down vote accepted

To prove that $X$ is closed, if $(x_0,y_0)$ is a limit point of $X$ that is not in $X$, consider the function $(x,y) \mapsto 1/f(x-x_0,y-y_0)$; this is just the reciprocal of the distance between the points $(x,y)$ and $(x_0,y_0)$.

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His $f$ doesn't mean the metric on $\mathbb R^2$, but the "distance function on $\mathbb R^2$", i.e. the norm. You got confused. Very unusual to call "$f$" a metric, but your argument works for the closed part. – Patrick Da Silva Nov 23 '13 at 6:30
@PatrickDaSilva Oops, I'll fix it. Thanks. – Trevor Wilson Nov 23 '13 at 6:31
Well now your $f$ is supposed to be his $f$, it's not the metric! I corrected it. – Patrick Da Silva Nov 23 '13 at 6:34
@PatrickDaSilva D'oh! Thanks for catching that. – Trevor Wilson Nov 23 '13 at 6:36
@ Mr. Wilson. Thank you for you hints. When $y \rightarrow x$, the function $\frac{1}{f(x,y)} \rightarrow \infty$. If $f(x,y) \neq 0$, $\frac{1}{f(x,y)}$ is continuous, that leads us to a contradiction. So $x$ must be in $X$. Hope it is right and the matter is clear to me. – Dutta Nov 23 '13 at 6:36

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