# Week 5 (418)

**tandr102**

Week 5

Research the separation axioms distinguishing *T*0, *T*1, *T*2, *T*3, and *T*4 spaces. Discuss some interesting aspects of these spaces to include their subtlety. Note that it is easy to create a space that is not even *T*0, e.g. *X* = {*a*,*b*,*c*,*d*,*e*} with Ƭ = {∅, *X*, {*a*,*b*,*d*},{*a*,*b*}, {*d*}} where the points *a* ∈ *X* and *b* ∈ *X* cannot be separated. As always comment on your classmates’ observations about the separation axioms, and remember to cite your work.

For example: A theorem regarding *T*1 spaces:

Theorem: A space *X* is a *T*1-space if and only if each finite subset of *X* is closed.

[Recall that *X* is T1 if, for any pair (x,y) of distinct elements of *X,* there are open sets *U* and *V* such that *U* contains x but not y and *V* contains y but not x.]

Proof: Suppose each finite subset of *X* is closed and consider distinct points *a*,*b* ∈ *X*. Then *U* = *X‒*{*a*} and *U* = *X‒*{*b*} are open sets since {*a*} and {*b*} are closed. Moreover, *U* contains *b* but not *a*, and *V* contains *a* but not *b*; thus *X* is *T*1. Now suppose *X* is *T*1. It is sufficient to prove that each singleton set {*a*} is closed since any finite set is the union of a finite number of singleton sets. To show {*a*} is closed, we show *U* = *X‒*{*a*} is open. For all *b* ∈ *X* where *b* ≠ *a* there is an open set *Ub* containing *b* but not *a* since *X* is *T*1. The union of the collection of sets *Ub* for all *b* ∈ *X* is open by the definition of a topology, but the union of all sets *Ub* is *X‒*{*a*}.

- 16 days ago
- 15

**Answer(1)**

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- ResearchtheseparationaxiomsdistinguishingT0.docx
- CamScanner10-12-202107.31.pdf

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