Every sequence in a compact space is a lingering sequence
- Note: this page is taking about a metric space not a topological space (at least at this time)
TODO: Does this apply at all to topological spaces? (Maybe, but with adaptation, as open balls do not exist in general top. spaces)
Contents
Statement
In a metric space [ilmath](X,d)[/ilmath] that is compact every sequence is a lingering sequence, that is to say^{[1]}:
- [math]\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][/math]
Note that such an [ilmath]x[/ilmath] exists, suggesting the space is complete. See Equivalent statements to compactness of a metric space for information
Proof
Suppose that it is untrue (this is a proof by contradiction), then we want to show that, given a [ilmath](x_n)_{n=1}^\infty\subseteq X[/ilmath] that:
- [ilmath]\forall x\in X\exists \epsilon > 0[\vert B_\epsilon(x)\cap(x_n)_{n=1}^\infty\vert\ne\aleph_0][/ilmath]
- Obviously as the intersection of sets is a subset of each set we see that [ilmath]B_\epsilon(x)\cap(x_n)_{n=1}^\infty\subseteq (x_n)_{n=1}^\infty[/ilmath] 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:
- [ilmath]\forall x\in X\exists \epsilon > 0[\vert B_\epsilon(x)\cap(x_n)_{n=1}^\infty\vert\in\mathbb{N}][/ilmath] we will assume this is true and reach a contradiction.
- Obviously as the intersection of sets is a subset of each set we see that [ilmath]B_\epsilon(x)\cap(x_n)_{n=1}^\infty\subseteq (x_n)_{n=1}^\infty[/ilmath] 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:
Proof:
- Let [ilmath](x_n)_{n=1}^\infty\subseteq X[/ilmath] be given
- Notice that for all [ilmath]x\in X[/ilmath] we get an associated [ilmath]\epsilon_x[/ilmath] such that the open ball [ilmath]B_{\epsilon_x}(x)[/ilmath] contains only finitely many terms of the sequence.
- Notice that this family, [ilmath]\{B_{\epsilon_x}(x)\}_{x\in X} [/ilmath] is actually an open cover of [ilmath]X[/ilmath] (a cover of [ilmath]X[/ilmath] by sets open in [ilmath](X,d)[/ilmath])
- However [ilmath]X[/ilmath] is compact by assumption, this means every open cover has a finite subcover. Thus:
- [ilmath]\exists\ \{x_1,\ldots,x_n\}\subseteq X[/ilmath] such that [ilmath]\{B_{\epsilon_{x_i} }(x_i)\}_{i=1}^n[/ilmath] is an open cover of [ilmath]X[/ilmath]
- If the [ilmath]n[/ilmath] balls in this cover each contain only finitely many terms of the sequence then their sum must be finite! That is:
- [math]\sum_{i=1}^n\left(\vert B_{\epsilon_{x_i} }(x)\cap(x_k)_{k=1}^\infty\vert\right)\in\mathbb{N}[/math] 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 [ilmath]X[/ilmath]
- This is obviously a contradiction, as there are countably many terms of the sequence in the space!
Thus we have shown if [ilmath]X[/ilmath] is compact then we cannot have [ilmath]\forall x\in X\exists \epsilon > 0[\vert B_\epsilon(x)\cap(x_n)_{n=1}^\infty\vert\in\mathbb{N}][/ilmath]. Thus we must have:
- [ilmath]\exists x\in X\forall\ \epsilon > 0[\vert B_\epsilon(x)\cap(x_n)_{n=1}^\infty\vert=\aleph_0][/ilmath]
Since the sequence was arbitrary, this is true for all/any sequences.
See also
References
- Todo
- Theorems
- Theorems, lemmas and corollaries
- Metric Space Theorems
- Metric Space Theorems, lemmas and corollaries
- Metric Space
- Real Analysis Theorems
- Real Analysis Theorems, lemmas and corollaries
- Real Analysis
- Functional Analysis Theorems
- Functional Analysis Theorems, lemmas and corollaries
- Functional Analysis