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Let X and Y be metric spaces and let f be a continuous function from X onto Y . Prove that if every sequence in X has a convergent subsequence then every sequence in Y has a convergent subsequence.

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closed as off-topic by Jack Lee, Daniel, drhab, MathOverview, Empty Oct 9 '15 at 16:34

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    $\begingroup$ Please, show some effort. Where are you stuck? $\endgroup$ – mfl Oct 8 '15 at 22:34
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Let $y_n$ be an arbitrary sequence in $Y$. Because $f$ is onto, so for every $n$ exists sequence $x_n$, that $f(x_n) = y_n$. From our assumption there exist convergent subsequence of $x_n$, say $x_{n_j}$. Consider subsequence $f(x_{n_j})$ of $f(x)$. I will show, that $f(x_{n_j})$ convergens to $f(x)$, where $x = \lim_{j\to\infty}x_{n_j}$. This folows from continuity of $f$. Let $\epsilon > 0$ be arbitrary. We can choose $\delta>0$, that $|f(x) -f(y)| < \epsilon$ holds for every $|x - y| < \delta$. There exist $N$, that $|x_{n_j}-x| < \delta$ holds for every $j > N$, so $|f(x_{n_j} - f(x)| < \epsilon$ holds for every $j > N$, so $\lim_{j\to\infty}f(x_{n_j}) = f(x)$.

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every sequence in $X$ has a convergent subsequence so $X$ is compact.

for every open cover of $Y$ by $f^{-1}$ we find an open cover of $X$ so it has a finite cover on $X$ and this cover show the subcover of $Y$ which is finite so $Y$ is compact and every sequnce in it has a convergent subsequence.

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  • $\begingroup$ Having shown $X$ is compact, why not just say $f(X)=Y$ is compact? $\endgroup$ – zhw. Oct 8 '15 at 23:22

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