Let $f:(X,d) \to (Y,d')$ be a continuous function and $(x_n)$ a convergent sequence in $X$. Show that $(f(x_n)) \in Y$ converges to $f(x')$.

Question

Let $f:(X,d) \to (Y,d')$ be a continuous function and $(x_n)$ a convergent sequence in $X$ that converges to $x'$. Show that $(f(x_n)) \in Y$ converges to $f(x')$.

Let $\varepsilon >0$. Since $f$ is continuous there exists $\delta > 0$ such that $f(B(x', \delta)) \subset B(f(x'), \varepsilon)$. Also since $x_n \to x'$ there exist $K \in \mathbb{N}$ such that $x_n \in B(x',\delta)$ for all $n > K$. Now what I would like to know that is it true that $f(x_n) \in B(f(x'), \varepsilon)$ whenever $n > K$? If this would be the case then it would follow immediately that $f(x_n) \to f(x')$, but I'm not sure this works?

Answer 1

$$x_n \in B(x',\delta) \implies f(x_n) \in f(B(x',\delta))$$ and $$f(B(x',\delta)) \subset B(f(x'),\epsilon)$$ So, $$x_n \in B(x',\delta) \implies f(x_n) \in B(f(x'),\epsilon)$$ So, yes, $f(x_n) \in B(f(x'),\epsilon)$ for $n>K$ and hence $f(x_n)\to f(x')$.