Problem 7C#Problem 7C (Hausdorff implies 𝑇1 axiom). Let 𝑋 be a Hausdorff topological space. Prove that for any point 𝑝∈𝑋 the set {𝑝} is closed.Solution by kiwiyouFor every 𝑞≠𝑝, there exists two open neighbourhoods 𝑈 of 𝑝 and 𝑉 of 𝑞 s.t. 𝑈∩𝑉=∅.Definition 7.3.1 (Hausdorff). A topological space 𝑋 is Hausdorff if for any two distinct points 𝑝,𝑞∈𝑋, there exists two open neighbourhoods 𝑈 of 𝑝 and 𝑉 of 𝑞 such that𝑈∩𝑉=∅.Let𝑌=⋃𝑞∈𝑋∖{𝑝}𝑉.Since 𝑌 is union of open neighbourhoods of 𝑝, 𝑌 is open.•Arbitrary unions (possibly infinite) of open sets are also open in 𝒯︀.Since 𝑌=𝑋∖{𝑝} is open, {𝑝} is closed.∎Definition 7.2.4. In a general topological space 𝑋, we say that 𝑆⊆𝑋 is closed in 𝑋 if the complement 𝑋∖𝑆 is open in 𝑋.
Problem 7E#Problem 7E. Let 𝑋 be a topological space. The connected component of a point 𝑝∈𝑋 is the union of all subspaces 𝑆⊆𝑋 which are connected and contain 𝑝.(a)Does the connected component of a point have to be itself connected?(b)Does the connected component of a point have to be an open subset of 𝑋?Solution by finalchild(a)Yes.Let 𝐶 be the connected component of 𝑝.Assume 𝐶=(𝐴∩𝐶)⊔(𝐵∩𝐶) where 𝐴∩𝐶 and 𝐵∩𝐶 are nonempty and 𝐴 and 𝐵 are open in 𝑋.Without loss of generality, assume 𝑝∈𝐵.Take some connected subspace 𝑆⊆𝑋 such that 𝐴∩𝑆≠∅.Then 𝑆=(𝐴∩𝑆)⊔(𝐵∩𝑆), and both 𝐴∩𝑆 and 𝐵∩𝑆 are nonempty and open in 𝑆.This is contradiction, so 𝐶 is connected. ∎(b)No.The connected components of 𝑝∈ℚ is not an open subset of ℚ. ∎