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Let $X$ be compact space and $f:X\rightarrow \mathbb{R}$ surjective function. Is $f$ continuous?

Im trying to prove it that it isn't continuous but I can't find the right example. So I tried:

Assume that $f$ is continuous. Then if $X$ is compact so $f(X)$ is. But since $f$ is surjective function, then $f(X)=\mathbb{R}$ and $\mathbb{R}$ is not compact. We get contradiction.

Am I thinking right? ( We consider $\mathbb{R}$ with natural topology)

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  • $\begingroup$ If you've covered that theorem, it's by far the easiest way to do it, yes. $\endgroup$ – Henno Brandsma Feb 2 '20 at 12:14
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You really only have one property of a continuous function: the preimage of an open set is an open set. You really only have one property of a compact set: any open cover is a finite subcover. This means we need to find some useful collection of open sets in $\Bbb{R}$ whose preimages are an open cover of $X$ and find a way that only finitely many of them cover $X$ contradicts surjectivity...

Suppose, for purpose of contradiction, $f$ is continuous. The preimage of every open set in $\Bbb{R}$ is open in $X$. For every $x \in X$, define the unit length open interval $$ U(x) = (f(x) - 1/2, f(x)+ 1/2) \subset \Bbb{R} $$ and note that $f(x) \in U(x)$. We have observed that $f^{-1}(U(x))$ is an open set in $X$. Since $f$ is a total function and each $x \in X$ satisfies $x \in f^{-1}(U(x))$, $\bigcup_{x \in X} f^{-1}(U(x))$ is an open cover of $X$. Since $X$ is compact this cover has a finite subcover. That is, there is an $S \subset X$ such that $|S|$ is finite and $\bigcup_{x \in S} f^{-1}(U(x)) = X$.

Since $f$ is surjective, $$ f(X) = f\left( \bigcup_{x \in S} f^{-1}(U(x)) \right) = \Bbb{R} \text{.} $$ But $f\left( \bigcup_{x \in S} f^{-1}(U(x)) \right)$ is a union of a finite collection of unit length open intervals of $\Bbb{R}$. Since $\Bbb{R}$ cannot be covered by a finite collection of unit length intervals, we have our contradiction.

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If $f$ is surjective, and $X$ is compact, then $f$ is not continuous: Pick $x_n \in X$ such that $f(x_n)=n$ for all $n \in \Bbb N$. (Does not even use the full force of surjectivity, btu it's all we need.)

As $X$ is compact, there is a point $p \in X$ such that for every open neighbourhood $O$ of $p$, $N(O):=\{n \in \Bbb N: x_n \in O\}$ is infinite. (This is called a point of complete accumulation of $\{x_n: n \in \Bbb N\}$.)

(Proof sketch : if no such $p$ exists, every $x \in X$ has an open neighbourhood $O_x$ with $N(O_x)$ finite. Take a finite subcover of $\{O_x: x \in X\}$ to derive a contradiction.)

Now take an open interval neighbourhood $I=(f(p)- \frac12, f(p)+\frac12)$ of $f(p)$. If $f$ were continuous at $p$ we'd have an open neighbourhood $O$ of $p$ such that $f[O] \subseteq I$, but then, for each $n \in N(O)$, $n = f(x_n) \in I$ and so $I$ contains infinitely many integers but is only of length $1$, which is absurd.

So we've found a point of non-continuity of $f$.

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