View source for Every sequence in a compact space is a lingering sequence/Proof

==Statement==
{{:Every sequence in a compact space is a lingering sequence/Statement}}
==Proof==
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Suppose that it is untrue (this is a proof by contradiction), then we want to show that, given a {{M|1=(x_n)_{n=1}^\infty\subseteq X}} that:
* {{M|1=\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 {{M|1=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:
*:* {{M|1=\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 {{M|1=(x_n)_{n=1}^\infty\subseteq X}} be given
# Notice that ''for all'' {{M|x\in X}} we get an associated {{M|\epsilon_x}} such that the open ball {{M|B_{\epsilon_x}(x)}} contains only finitely many terms of the sequence.
# Notice that this family, {{M|\{B_{\epsilon_x}(x)\}_{x\in X} }} is actually an [[open cover]] of {{M|X}} (a [[cover]] of {{M|X}} by sets [[open set|open]] in {{M|(X,d)}})
# However {{M|X}} is [[compact]] by assumption, this means every open cover has a finite subcover. Thus:
#* {{M|\exists\ \{x_1,\ldots,x_n\}\subseteq X}} such that {{M|1=\{B_{\epsilon_{x_i} }(x_i)\}_{i=1}^n}} is an open cover of {{M|X}}
# If the {{M|n}} balls in this cover each contain only finitely many terms of the sequence then their sum must be finite! That is:
#* {{MM|1=\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.
# But the family of balls cover the space! This suggests there is at most a finite number of terms of the sequence in {{M|X}} 
# This is obviously a contradiction, as there are countably many terms of the sequence in the space!
Thus we have shown if {{M|X}} is compact then we ''cannot have'' {{M|1=\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:
* {{M|1=\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}}
==References==
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