Question

If 𝑓 is a continuous function from a metric space X into a metric space Y and {𝑥𝑛} is a sequence in X

which converges to 𝑥 then show that the sequence {𝑓(𝑥𝑛)} converges to 𝑓(𝑥) in Y.

Answer 1

𝑓 is a continuous function from a metric space X into a metric space Y if and only if $f$ is continuous at every point of X.

$f(x)$ is the limit of the sequence $f(x_n)$ if and only if

(1) $\forall \varepsilon>0\, \exists N\in\mathbb{N}\, \forall n>N\, d_Y(f(x),f(x_n))<\varepsilon$

Fix $\varepsilon>0$. Since 𝑓 is a continuous function at point $x\in X$ then

(2) $\exists\delta>0\, \forall x': d_X(x,x')<\delta\to d_Y(f(x),f(x'))<\varepsilon$

Since $x=\lim\limits_{n\to\infty}x_n$ , there exists $N\in\N$ such that for all $n>N$$d_X(x,x')<\delta$.

Applying (2), we get $d_Y(f(x),f(x_n))<\varepsilon$.

Finally, we have deduced (1), and the assertion is proved.