1
$\begingroup$

Let $f:X\rightarrow Y$ be a continuous map of metric spaces. Show that if $A\subseteq X$ is compact, then $f(A)\subseteq Y$ is compact.

I am using this theorem: If $A\subseteq X$ is sequentially compact, it is compact. Also this definition: A set $A\subseteq X$ is sequentially compact if every sequence in $A$ has a convergent subsequence in $A$.

Attempt at a proof:

Let $\{y_n\}\subseteq f(A)$. Since $f$ is continuous, $\{y_n\}=f(x_n)$ for some $\{x_n\}\subseteq A$. If $A\subseteq X$ is compact, every sequence $\{x_n\}\subseteq A$ has a subsequence that converges to a point in $A$, say $\{x_{n_k}\}\rightarrow a\in A$. Since $f$ is continuous, $f(x_{n_k})\rightarrow f(a)\in f(A)$. Then $f(x_{n_k})\subseteq f(A)$ is a convergent sequence in $f(A)\implies f(A)$ is compact since $\{y_n\}\subseteq f(A)$ was arbitrary.

  • 2
    Almost right; formally, you need to start by taking an arbitrary sequence in $f(A)$ and show that it has a convergent subsequence; but such a sequence is clearly of the form $f(x_{n})$ for some sequence $\{x_n\}$ in $A$. Compactness of $A$ will get you a convergent subsequence $x_{n_k}$. As you correctly note, $f(x_{n_k})$ is then convergent in $f(A)$, which proves the result.2012-02-12
  • 0
    I'm still unsure how having a sequence in $f(A)$ immediately implies that it is a sequence $f(x_n)$ for a sequence $\{x_n\}\subseteq A$. Is it because $f$ is continuous?2012-02-12
  • 0
    $f(A)$ is the set of points $f(x)$ for $x\in A$...2012-02-12

4 Answers 4