Proposition

Suppose $X$ is a topological space, $A$ is a subset of $X$, and $(x_i)$ is a sequence of points in $A$ that converges to a point $x \in X$. Show that $x \in \overline{A}$.

Solution

Suppose $x \notin \overline{A}$.

Then $\overline{A}^c$ is a neighborhood of $x$. Since $x_i \rightarrow x$, there must exist an $N \in \mathbb{N}$ such that $\forall n \geq N, x_n \in \overline{A}^c$. Since all the points of the sequence are in $A$, they are not in $\overline{A}^c$. Therefore, no such $N$ exists.

This implies that $x$ must be in $\overline{A}$.