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Let $X\subset \mathbb R^{n}$ be a compact set, and $f :\mathbb R^{n}\to \mathbb R $ a continuous function. Then, $F(X)$ is a compact set.

I know that this question may be a duplicate, but the problem is that I have to prove this using real analysis instead of topology.

I'm struggling with proving that $F(X)$ is bounded. I know that the image of a continuous function is bounded, but I'm having trouble when it comes to prove this for vectorial functions.

If somebody could help me with a step-to-step proof, that would be great.

Peter_Pan
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Lessa121
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  • Do you know how to prove that a continuous function on a closed set is bounded? – Nishant Jul 21 '14 at 19:43
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    I think if OP knew that, she could have showed that continuous function on a compact set is bounded. She is already using the property that compact sets are closed and bounded. – Juanito Jul 21 '14 at 19:44

3 Answers3

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Take any open cover of F(K), as F is continuous, the inverse images of those open sets form an open cover of K. Since K is compact there is a finite subcover. By construction, the images of the finite subcover give a finite subcover of F(K), therefore F(K) is compact.

Random Excess
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Let any sequence $(y_n) \in F(K)^{\mathbb N}$ that converges to some $y$. Then by definition there is some $(x_n) \in K^{\mathbb N}$ such that $F(x_n)=y_n$ for every $n$. Since $K$ is compact (and thus closed and bounded), there is a subsequence $(x_{n_k})$ of $(x_n)$ that converges to some $x\in K$ (see Bolzano Weierstrass Theorem). By continuity of $F$ we have $$ y = \lim_{k \to \infty}y_{n_k} = \lim_{k \to \infty}F(x_{n_k}) = F(x).$$ It follows that $y \in F(K),$ and thus $F(K)$ is closed.

Now suppose that there is a sequence $(w_n) \in F(K)^{\mathbb N}$ such that $w_n \geq n$ for every $n \in \mathbb{N}$. Again for every $n$ there is $v_n \in K$ such that $F(v_n)=w_n$. $(v_n)$ is contained in the compact $K$ and thus admits a subsequence $v_{n_k}$ converging to some $v \in K$. But this is not possible since $F(v) \in \mathbb{R}$ and $$\infty = \lim_{k \to \infty }n_k \leq \lim_{k \to \infty } F(v_{n_k})=F(v).$$ A contradiction, thus $F(K)$ must be bounded.

It finally follows that $F(K)$ is compact.

Surb
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Suppose that f(X) isn't bounded, then it means that f tend to infinity somewhere on X (because X is compact), so f isn't continuous on one point of X and you have a contradiction.

Try to write it clearly, of course, this is just a "hint"

nicomezi
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