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Assume $f:X\rightarrow Y$, where $X$ and $Y$ are two metric spaces. If $f(\overline{E})\subset \overline{f(E)}, \, \forall E\subset X$, then how can we prove that $f$ is continuous?

Thank you for your advice.

User 1e5
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Since $X$ and $Y$ are metric spaces, continuity is equivalent to sequential continuity.

Let $(x_n)_n$ be a sequence in $X$ which converges to some $x\in X$. We have to show that $f(x_n)$ converges to $f(x)$ in $Y$.

It suffices to show that every subsequence $(x_{n_k})_k$ has a further subsequence $(x_{n_{k_i}})_i$ such that $f(x_{n_{k_i}})$ converges to $f(x)$ as $i\rightarrow\infty$ (see for example here).

Let $(x_{n_k})_k$ be a subsequence and $E=\{x_{n_k}\,:\,k\in\mathbb{N}\}$. Then $\overline{E}=E\cup\{x\}$. Therefore $f(x)\in f(\overline{E})$ and by assumption, $f(x)\in \overline{f(E)}$. Thus there exists a sequence in $f(E)=\{f(x_{n_k})\,:\,k\in\mathbb{N}\}$ that converges to $f(x)$. That is, a subsequence of the chosen subsequence converges to $f(x)$.

Therefore $f(x_n)$ converges to $f(x)$ and $f$ is continuous.

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J.R.
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  • Thank you for your response. It is insightful. Also, could you guide me on the proof that if for every convergent sequence ${p_n}$ in $X$, ${f(p_n)}$ converges in $Y$, then $f(\overline{E})\subset \overline{f(E)}$? thank you. – User 1e5 Apr 11 '14 at 16:24