Every Inf. Subset of a Comp. Set in a Met. Space Contains an Acc. Pt.

# Every Infinite Subset of a Compact Set in a Metric Space Contains an Accumulation Point

Recall from the Compact Sets in a Metric Space page that if $(M, d)$ is a metric space then a subset $S \subseteq M$ is said to be compact if every open covering of $S$ contains a finite subcovering.

On the Boundedness of Compact Sets in a Metric Space page that if $(M, d)$ is a metric space and $S \subseteq M$ is a compact subset of $M$ then $S$ is also a bounded set in $M$.

We also saw on the Closedness of Compact Sets in a Metric Space page that if $S$ is a compact subset of $M$ then furthermore $S$ is also closed in $M$.

We will now look at another nice property that arises from compact sets - namely that every infinite subset of a compact set contains an accumulation point.

Theorem 1: Let $(M, d)$ be a metric space and let $S \subseteq M$. If $S$ is a compact then every infinite subset $X \subseteq S$ contains an accumulation point. |

**Proof:**Suppose that $S$ is a compact set. Let $X \subseteq S$ be any infinite subset of $S$ and assume that $X$ does not contain any accumulation points. We will show that this results in a contradiction.

- Since $X$ does not contain any accumulation points we must have that for all $x \in M$ that there exists positive real numbers $r_x > 0$ such that the ball centered at $x$ with radius $r_x$ does not contain any points of $S$ different from $x$, that is, for all $x \in M$:

\begin{align} \quad B(x, r_x) \cap S \setminus \{ x \} = \emptyset \end{align}

- For each $x \in S$, consider the collection of open balls $\{ B(x, r_x) : x \in S \}$. Then we have that:

\begin{align} S \subseteq \bigcup_{x \in S} B(x, r_x) \end{align}

- This is an infinite open covering of $S$, however, no such finite open subcovering exists because each element in this infinite open covering of $S$ contains only one point of $S$ (namely $x$). This implies that $S$ is not compact which is a contradiction. Therefore the assumption that the infinite subset $X$ of $S$ does not have any accumulation points is false. Hence every infinite subset $X \subseteq S$ has an accumulation point.