Every sequence in a compact space is a lingering sequence/Proof

Statement

In a metric space $(X,d)$ that is compact every sequence is a lingering sequence, that is to say:

• $$\forall(x_n)_{n=1}^\infty\subseteq X\ :\ \exists x\in X\ \forall\epsilon>0[\vert B_\epsilon(x)\cap(x_n)_{n=1}^\infty\vert=\aleph_0]$$

Proof

Suppose that it is untrue (this is a proof by contradiction), then we want to show that, given a $(x_n)_{n=1}^\infty\subseteq X$ that:

• $\forall x\in X\exists \epsilon > 0[\vert B_\epsilon(x)\cap(x_n)_{n=1}^\infty\vert\ne\aleph_0]$

  Obviously as the intersection of sets is a subset of each set we see that $B_\epsilon(x)\cap(x_n)_{n=1}^\infty\subseteq (x_n)_{n=1}^\infty$ and a sequence contains a countably infinite amount of terms. As a subset has cardinality less than or equal to a set we see that $\ne\aleph_0$ means showing that the intersection is finite, or:

  • $\forall x\in X\exists \epsilon > 0[\vert B_\epsilon(x)\cap(x_n)_{n=1}^\infty\vert\in\mathbb{N}]$ we will assume this is true and reach a contradiction.

Proof:

• Let $(x_n)_{n=1}^\infty\subseteq X$ be given

1. Notice that for all $x\in X$ we get an associated $\epsilon_x$ such that the open ball $B_{\epsilon_x}(x)$ contains only finitely many terms of the sequence.
2. Notice that this family, $\{B_{\epsilon_x}(x)\}_{x\in X}$ is actually an open cover of $X$ (a cover of $X$ by sets open in $(X,d)$)
3. However $X$ is compact by assumption, this means every open cover has a finite subcover. Thus:
   • $\exists\ \{x_1,\ldots,x_n\}\subseteq X$ such that $\{B_{\epsilon_{x_i} }(x_i)\}_{i=1}^n$ is an open cover of $X$
4. If the $n$ balls in this cover each contain only finitely many terms of the sequence then their sum must be finite! That is:
   • $$\sum_{i=1}^n\left(\vert B_{\epsilon_{x_i} }(x)\cap(x_k)_{k=1}^\infty\vert\right)\in\mathbb{N}$$ is a finite sum of finite numbers, thus is finite.
5. But the family of balls cover the space! This suggests there is at most a finite number of terms of the sequence in $X$
6. This is obviously a contradiction, as there are countably many terms of the sequence in the space!

Thus we have shown if $X$ is compact then we cannot have $\forall x\in X\exists \epsilon > 0[\vert B_\epsilon(x)\cap(x_n)_{n=1}^\infty\vert\in\mathbb{N}]$. Thus we must have:

• $\exists x\in X\forall\ \epsilon > 0[\vert B_\epsilon(x)\cap(x_n)_{n=1}^\infty\vert=\aleph_0]$

Since the sequence was arbitrary, this is true for all/any sequences.

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TODO: Make this more formal with cardinality arguments

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