Question

I literally in a desperate situation trying to prove the converse to the theorem that every continuous real valued function on a compact metric space achieves a maximum and a minimum value.

The problem has been posted below, I sincerely hope any expert could give me a hand, Thanks so so much!!!

Let (X, d) be a metric space. Show that if every continuous real valued function on X achieves maximum and minimum values, then X must be compact.

Answer 1

( X , d) be a metric space .

Suppose every continuous real valued function on X achives maximum and minimum values .

X is not bounded , then consider the continuois map ,

f : X $\rightarrow$ R be defined by f(x) = ||x || .

As X is not bounded so f is also not bounded , a contradiction to our assumption . So X is bounded.

Also if X is not closed the there exist a limit point a of X which does not belongs to X .

As a is limit point so for each $\epsilon > 0$ there exist x such that

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|| x - a ||​​​$< \epsilon$

Now consider the map f : X $\rightarrow$ R be defined by ,

1 f(x) E-a

Which is well defined as a is not belongs to X. Then f is continuous but not bounded , a contradiction to our assumption.  

So X is closed.

I.e., f is closed as well as f is bounded .

Hence f is compact.

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If you have any doubt or need more clarification at any step please comment .