Convergence of a sequence in relative topology

Question

Let X be a topological space X. Show that sequence {${x_i}$} in S converges to $x_0$ in S in relative topology if and only if the sequence {${x_i}$} converges to $x_0$. Can anyone help me out with this problem. For the reverse direction would I first assume that if the sequence {${x_i}$} does not converges to $x_0$ to prove the sequence convergence in S? The forward direction makes sense, but I am not sure how to show it.

Answer 1

$i: S \to X$ defined by $i(x) = x$ is (by definition of the subspace topology) continuous, so if $x_n \in S$ for all $S$ and $x_n \to x_0$ in $S$, then $i(x_n) =x_n \to i(x_0) = x_0$ in $X$.

And if $x_n \to x$ in $X$ and $O \cap S$ is an open neighbourhood of $x_0$, then $x_0 \in O$ in particular, so $x_n \in O$ for almost all $n$, hence $x_n \in O \cap S$ for almost all $n$, and so $x_n \to x_0$ in $S$.