Let $A\subset \Bbb R^2 $ with the property that every continuous function on $A$ has a maximum in $A$ .Prove that $A$ is compact.

Question

My try:

We have to show that $A$ is closed and bounded.

In order to prove this result we should have some ready continuous functions in our hand. The projection maps can be used.

Define $\pi_x:A\to \Bbb R ;\pi_x(x,y)=x$ and $\pi_y:A\to \Bbb R ;\pi_y(x,y)=y$

They attain their bounds on $A$ so there exists $(a,b)\in A$ such that $\sup_{\{(x,y)\in A\}}\pi _x(x,y)=a $ and $(a,b)\in A$ such that $\sup_{\{(x,y)\in A\}}\pi _y(x,y)=b$

I can't proceed further.Please help

Possible duplicate of http://math.stackexchange.com/questions/668905/if-every-real-valued-continuous-function-is-bounded-on-x-metric-space-then — detnvvp, Mar 17 '16 at 10:14
@detnvvp I would vote for this question to remain, because the one you linked to is a generalization of this one (this one is focused on $\mathbb R^2$ and may be easier to understand for less experienced users) — 5xum, Mar 17 '16 at 10:14
Yes, I also agree; just to note that this holds in a more general setting. — detnvvp, Mar 17 '16 at 10:16
Is the answer you recieved what you wanted? If yes, you should accept it. If no, you should explain what is missing. — 5xum, Mar 18 '16 at 10:34

score 4 · Accepted Answer · answered Mar 17 '16 at 10:12

You can prove that $A$ is bounded by contradiction. Assume that $A$ is not bounded, and look at the function $f(x,y)=x^2+y^2$. Does this function have a maximum?

Similarly, you can prove that $A$ is closed. Assume that it is not. That means there exists a point $(x_0, y_0)$ which is in the closure of $A$, but not in $A$. Now, look at the function $f(x,y)=\frac{1}{(x-x_0)^2 + (y-y_0)^2}$

Let $A\subset \Bbb R^2 $ with the property that every continuous function on $A$ has a maximum in $A$ .Prove that $A$ is compact.

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