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An exercise from Munkres states that:

Suppose that $f: X \to Y$ is continuous. If $x$ is a limit point of a subset of $A$ of $X$, is it necessarily true that $f(x)$ is a limit point of $f(A)$?

I think that the answer is no based on an example constant function, but I am curious as to how to fix the statement so it becomes true. For instance, is this a correct statement, or do I need to add more conditions on $f$ or $A$?

Suppose that $f: X \to Y$ is continuous. If $x$ is a limit point of a subset of $A$ of $X$ such that $f(A)$ has limit points, then $f(x)$ is a limit point of $f(A)$.

user96608
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    What do you mean by "limit point"? If it is an element of the closure, then the statement looks true to me. – Giuseppe Negro Sep 24 '13 at 14:30
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    @Giuseppe: The OP likely means "not an isolated point." – Cameron Buie Sep 24 '13 at 14:30
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    You are correct. Your counterexample, a constant function, is a good one. What additional condition could you impose on $f$, to make it as far from being a constant function as possible, that might make the statement true? – Brett Frankel Sep 24 '13 at 14:49

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There is an equivalent condition to being continuous: $f[\overline{A}] ⊆ \overline{f[A]}$ for all $A ⊆ X$. If $x$ is a limit point of $A$ then $x ∈ \overline{A}$ so $f(x) ∈ \overline{f[A]}$. If $f(x) ∈ \overline{f[A]} \setminus f[A]$ then it is a limit point of $f[A]$. So if $f[A]$ has no isolated points then it holds.

It is not enought to assume that $f[A]$ has limit points since you can always take a topological sum of identity on non-discrete space (so $f[A]$ can contain a limit point) and a constant function (so the proposition doesn't hold).

user87690
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