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Let $(X, \mathcal{T}) and (Y, \mathcal{U})$ be topological spaces and let $f : X \rightarrow Y$.

Suppose $f$ is continuous, and $\{x_n\}_{n=1}^{\infty}$ is a sequence in $X$ converging to a point $x$.

I need to show that $\{f(x_n)\}_{n=1}^{\infty}$ converges to $f(x)$.

This is what I have so far:

Since $f$ is continuous, we have that $f(\overline A) =$ $\overline{f(A)}$. We see that $\overline{\{x_n\}_{n=1}^{\infty}} =$ $\{x_n\}_{n=1}^{\infty} \cup \{x\}$ and so we have $f(\{x_n\}_{n=1}^{\infty} \cup \{x\}) =$ $\overline{\{f(x_n)\}_{n=1}^{\infty}} = \{f(x_n)\}_{n=1}^{\infty} \cup$ $f(x)$. Therefore, there exists a sequence in $\{f(x_n)\}_{n=1}^{\infty}$ that converges to $f(x)$.

But now I'm stuck because I don't know how to show that this sequence is $\{f(x_n)\}_{n=1}^{\infty}$ and not just some subsequence of it?

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1 Answer 1

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For any neighborhood $V$ of $f(x)$, $f^{-1}(V)$ is a neighborhood of $x$, so there exists some $N\in \mathbb{N}$ such that $x_n\in f^{-1}(V) \forall n>N,$ $i.e. f(x_n)\in V\forall n>\mathbb{N}$.

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