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Without using the idea of compact or sequential compactness , can we prove that if $A$ is a closed bounded set in $\mathbb R^2$ , then any continuous function $f:A \to \mathbb R$ is bounded and attains it's bounds ?

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    $\begingroup$ Since this is wrong in infinite dimensional spaces, probably not. Somewhere in the proof you need something that's equivalent to compactness. $\endgroup$ – user251257 Jul 15 '15 at 15:15
  • $\begingroup$ How can we prove a topological property of the range of some functions without using topological properties of the underlying space? $\endgroup$ – Jack D'Aurizio Jul 15 '15 at 15:17
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    $\begingroup$ Can you use the fact that it's true for functions on closed bounded sets in $\mathbb R$? $\endgroup$ – Robert Israel Jul 15 '15 at 15:17
  • $\begingroup$ @RobertIsrael : Yes , yes we can use that it's true for functions on closed bounded sets in $\mathbb R$ $\endgroup$ – user228168 Jul 15 '15 at 15:21
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Since $A$ is bounded, take $R$ so $d(0,p) \le R$ for all $p \in A$, where $d$ is Euclidean distance. In particular, the $x$ and $y$ coordinates of each point of $A$ are in $[-R,R]$.

Let $J = \pi(A) \subset \mathbb R$ be the image of $A$ under the projection $\pi: (x,y) \to x$ of $\mathbb R^2$ onto $\mathbb R$. I claim this is closed. To show this, suppose $x_0$ is in the closure of $J$. For any $y$, let $e(y)$ be the distance from $(x_0,y)$ to $A$, i.e. $\inf \{d((x_0,y), p): p \in A\}$. Using the triangle inequality, $|e(y) - e(y')| \le |y - y'|$, so $e$ is continuous, and therefore $e$ attains a minimum on the closed, bounded interval $[-R, R]$. But that minimum must be $0$, because for any $\epsilon > 0$ there is $(x,y) \in A$ with $|x - x_0| < \epsilon$, making $e(y) \le d((x_0,y), (x,y)) = |x - x_0| < \epsilon$. If $e(y) = 0$, $(x_0,y)$ is in the closure of $A$, and therefore in $A$, so $x$ is in $J$.

Let $F(p) = \arctan(f(p))$. Thus $F: A \to \mathbb R$ is continuous and bounded ($|F(p)| < \pi/2$). It suffices to show that $F$ attains its supremum on $A$: if $\sup \{F(p): p \in A\} = F(q)$, then $f(q) = \tan(F(q)) = \sup\{f(p): p \in A\}$ (and in particular $f$ is bounded above); similarly for the infimum, replacing $F$ by $-F$.

Let $s = \sup \{F(p): p \in A\}$. For each positive integer $n$, let $A_n = \{p \in A: F(p) \ge s - 1/n\}$, and $J_n = \pi(A_n)$. Then each $A_n$ is closed and bounded, and so by the previous result $J_n$ is closed and bounded. Moreover, $J_n$ is nonempty.

Define functions $g_n$ on $J$ so that $g_n(x)$ is the distance from $x$ to $J_n$, i.e. $$g_n(x) = \inf \{|x - z|: z \in J_n\}$$ Then $g_n$ is continuous and bounded on $J$, with $|g_n(x) - g_n(x')| \le |x - x'|$. Let $$g(x) = \sum_{n=1}^\infty 2^{-n} g_n(x)$$ Then $g$ is continuous and bounded on $J$. Now if $x \in J_m$, $g_n(x) = 0$ for all $n \le m$, so $g(x) \le \sum_{n=m+1}^\infty 2^{-n} (2 R) \le 2^{1-m} R$. Thus the infimum of $g$ on $J$ is $0$. By the one-dimensional result, $g$ attains its infimum, i.e. there is some $x_0$ such that $g(x_0) = 0$, and this means $x_0 \in J_n$ for all $n$. That says for every $n$ there is $y$ with $(x_0,y) \in A$ and $F(x_0,y) \ge s - 1/n$.

Now let $K = \{y: (x_0, y) \in A\}$, which is a closed bounded set, and on it define $h(y) = F(x_0, y)$, which is a continuous function. By the preceding paragraph we see that $\sup \{h(y): y \in K\} = s$. Again using the one-dimensional result, $h$ attains its supremum, i.e. there is some $y_0$ such that $F(x_0, y_0) = s$. And this is what we needed to show.

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  • $\begingroup$ That the intersection of $J_n$ isn't empty, looks suspiciously like Cantor's intersection theorem. That can be used to prove Heine-Borel ... :) $\endgroup$ – user251257 Jul 15 '15 at 21:54

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