Prove that a closed set contains all its limit points

Question

The following is the definition of an open set from Rudin's Real and Complex Analysis (3rd edition):

A closed set would be an open set's complement. From this definition, why would a closed set contain its limit points?

Answer 1

You have to use the topological notion of a limit: A sequence $(x_{n})$ in $X$ converges to some $x\in X$ if and only if for all open sets $V\subset X$ containing $x$ there exists an $N\in\mathbb{N}$ such that for all $n>N$ we have $x_{n}\in V$.

Now let $C$ a closed set, suppose there is some sequence $(x_{n})$ in $C$ converging to some $x\in X\setminus C$. By definition $X\setminus C$ is an open set so there exists an $N\in \mathbb{N}$ such that for all $n>N$ we have $x_{n}\in X\setminus C$, which is a contradiction.

Answer 2

Let $X$ be a topological space and $S$ a closed subset of $X$.

If $x$ is a limit point of $S$, then each open set $U$ with $x \in U$ has the property that $ U \cap (S \setminus \{x\}) \ne \emptyset.$

Now suppose that $x \notin S$. Then $U:= X \setminus S $ is open and $x \in U.$

Then we have $ U \cap (S \setminus \{x\}) \ne \emptyset$ , on the other hand we have $ U \cap (S \setminus \{x\}) = \emptyset$, a contradiction.

So: $x \in S.$