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Could you give me a hint on this problem?

Show that $f:A\subset\mathbb{R}^n\longrightarrow \mathbb{R}^m$ is continuous if and only if for every subset $B$ of $A$ , $f(A\cap\overline{B}) \subset \overline{f(B)}$.

Currently I know these definitions of continuity:

$(\rm i)$ In terms of pre-image of open sets.

$(\rm ii)$ In terms of $\epsilon$-$\delta$

$(\rm iii)$ In terms of convergent sequences.

By the statement of the problem I could guess the definition to use in this case is the first one. Maybe you could tell me how to start so I can give it a try, thanks in advance.

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Would you mind telling us your definition of continuity? Just so we're starting at the same place. –  Dylan Moreland Jul 18 '12 at 23:53
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2 Answers

up vote 2 down vote accepted

HINTS:

($\Rightarrow$) Let $x\in A\cap\operatorname{cl}B$. There is a sequence $\langle x_n:n\in\Bbb N\rangle$ in $B$ that converges to $x$. What do you know about $\langle f(x_n):n\in\Bbb N\rangle$?

($\Leftarrow$) Prove the contrapositive: suppose that $f$ is not continuous, and show that there is some $B\subseteq A$ such that $f[A\cap\operatorname{cl}B]\nsubseteq\operatorname{cl}f[B]$. If $f$ is not continuous, there is at least one point $x\in A$ such that $f$ is not continuous at $x$. Can you show that there are an $\epsilon>0$ and a sequence $\langle x_n:n\in\Bbb N\rangle$ in $A$ that converges to $x$ such that $\|f(x_n)-f(x)\|\ge\epsilon$ for every $n\in\Bbb N$? What can you then say about $B=\{x_n:n\in\Bbb N\}$?

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To do (⇐) I think it's a bit slicker (and doesn't use the metric structure) to take B to be the preimage of an arbitary closed set in the range. What does the condition tell you?

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