Suppose that \(f: X \to Y\) is continuous. If \(x\) is a limit point of the subset \(A\) of \(X\), is it necessarily true that \(f(x)\) is a limit point of \(f(A)\)?
Solution
1. No: for the constant map \(f\equiv0\colon\RR\to\RR\) and \(A=\theset{1/n : n\in\NN}\), the point \(0\) is a limit point of \(A\), but \(f(0)\) is not a limit point of \(f(A)\).
A constant map is continuous. Every interval about \(0\) contains \(1/n\ne0\) for large \(n\), so \(0\) is a limit point of \(A\). But \(f(A)=\theset{0}\), and a limit point of \(\theset{0}\) would need neighborhoods meeting \(\theset{0}\) in a point other than itself, which is impossible for \(f(0)=0\).
2. For continuous \(f\) and a limit point \(x\) of \(A\), \(f(x)\in\overline{f(A)}\); if moreover \(f(x)\notin f(A)\), then \(f(x)\) is a limit point of \(f(A)\).
Let \(V\) be a neighborhood of \(f(x)\). Then \(f^{-1}(V)\) is a neighborhood of \(x\), so it contains a point \(a\in A\) with \(a\ne x\), and \(f(a)\in V\cap f(A)\). If \(f(x)\notin f(A)\), the point \(f(a)\) differs from \(f(x)\).
3. Q.E.D.
Step 1 answers the question.