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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.

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    $\begingroup$ 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. $\endgroup$ – Martin Wanvik Feb 12 '12 at 3:05
  • $\begingroup$ 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? $\endgroup$ – Emir Feb 12 '12 at 3:30
  • $\begingroup$ $f(A)$ is the set of points $f(x)$ for $x\in A$... $\endgroup$ – David Mitra Feb 12 '12 at 3:31
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Actually the theorem is true for general topological spaces. Suppose $\{C_\alpha\}$ is an open cover of $f(A)$. The preimage under $f$ of each $C_\alpha$, $f^{-1}(C_\alpha)$, is open in $X$ since $f$ is continuous. Moreover, since $f(A)\subset\bigcup_\alpha C_\alpha$, we have $A\subset\bigcup_\alpha f^{-1}(C_\alpha)$. Since $A$ is compact, there exist finitely many $\alpha_1,\dots,\alpha_n$, such that $A\subset\bigcup_{i = 1}^n f^{-1}(C_{\alpha_i})$. Therefore, $f(A)\subset \bigcup_{i = 1}^n C_{\alpha_i}$. That is, every open cover of $f(A)$ admits a finite subcover.

The only catch here is that the preimage of an open set under $f$ is open. This is the definition of continuity in general topological spaces, and follows easily from the "standard" definition of continuity in metric spaces (by standard I mean $\epsilon$-$\delta$). Try to prove it.

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You need to start with a sequence $\{y_n\}\subset f(A)$ and prove that it has a convergent subsequence. (you can use the same ideas you were trying before though)

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Just writing the standard proof in a verbal (easy) form. Take an open covering of $f(A)$ in $Y$. We want to produce a finite subcover for this covering. Consider the inverse image of the elements of this covering under $f$. Note that the continuity of $f$ imply that these inverse images are open in $X$. Also they form an open covering of $A$. This Covering of $A$ has a finite subcover. Now consider the image of elements in this finite cover, they form the required sub-cover.

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Yet another formulation for topological spaces: If $f:X \to Y$ continuous and $f(x_\iota)$ is a net in $f(X)$, then $x_\iota$ has a converging subnet, say $x_\tau \to x$. Then $f(x_\tau) \to f(x)$, hence $f(x_\iota)$ has a converging subnet, so $f(X)$ is compact.

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