Characterization of continuity via closed.

Question

Let $K \subset\mathbb R^{n}$ a compact set and $f : K \rightarrow\mathbb R^{m}$ a continuous and one-to-one function. Show that the function $f^{-1} : f(K) \rightarrow K$ it is continuous.

Hint: By hip $K$ is compact and f is injective, so $f(K)$ is compact, then by the Heine-Borel theorem is closed and bounded.

It is just another way of saying $f^{-1}$ is continuous in this context. — Arctic Char, Jun 19 '21 at 03:23
@JoséRodríguez In particular, a homeomorphism is a continuous bijection with a continuous inverse. And yes, the easiest way to show the inverse is continuous is with the "Compact to hausdorff" theorem linked above — Alan, Jun 19 '21 at 04:15
Let $g: f[K] \to K$ be the inverse (which exists as $f: K \to f[K]$ is a bijection). If $C \subseteq K$ is closed, what is $g^{-1}[C]$ equal to? — Henno Brandsma, Jun 19 '21 at 07:12

score 1 · Answer 1 · answered Jun 20 '21 at 04:10

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To show $f^{-1}$ continuous it's enough to see that pre-image of any closed set in $K$ under $f^{-1}$ is closed in $f(K)$.

Let $V$ be any closed set in $K$. Then $V$ is compact (can be proved easily). As you said $f(V)$ is compact in $f(K)$ (continuous image). By Heine-Borel theorem $f(V)$ is then closed in $f(K)$. Since $(f^{-1})^{-1}(V)=f(V)$. We're done.

answered Jun 20 '21 at 04:10

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Thanks! I will analyze it! – José Rodríguez Jun 22 '21 at 20:04

Characterization of continuity via closed.

1 Answers1