Hausdorff spaces

“Non-Hausdorff spaces don’t exist in nature”

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Overview §

The key difference between sequences in metric spaces and sequences in arbitrary topological spaces is that a sequence in arbitrary space can converge to more than one point.

A sequence $x_1,x_2,\dots$ of points in an arbitrary topological space $X$ converges to the point $x$ if, for each neighborhood $U$ of $x$, there exists a positive integer $N$ such that $x_n\in U$ for all $n\ge N$.

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Hausdorff spaces impose an additional constraint on an arbitrary space so that sequences converge to one point only (i.e., one-point sets are closed).

Definition: Hausdorff space

A topological space $X$ is called a Hausdorff space if for each pair of distinct points $x_1,x_2\in X$, there exist neighborhoods $U_1,U_2$ of $x_1,x_2$, respectively, such that $U_1\cap U_2=\varnothing$.

Theorem (Munkres 17.8 & 17.10)

If $X$ is Hausdorff, then:

• (a) Every finite subset of $X$ is closed;
• (b) Every sequence in $X$ converges to at most one point of $X$.

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The $T_1$ axiom §

Definition: The T_1 axiom

The $T_1$ axiom is the condition that all finite subsets of a space are closed.

Note that this is weaker than the Hausdorff axiom!

Proposition (Munkres 17.9)

Let $X$ be a topological space that satisfies the $T_1$ axiom (i.e., all of its finite subsets are closed) and $A\subset X$ be any subset. Then a point $x\in A$ is a limit point of $A$ if and only if every neighborhood of $x$ contains infinitely many points of $A$.

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Review §

• Show that the topological definition of a convergent sequence agrees with the usual $ϵ-\delta$ definition of convergence in $\mathbb{R}$. ⭐
• Prove the theorem that if a set $X$ is Hausdorff, then (a) every finite subset of $X$ is closed and (b) every sequence in $X$ converges to at most one point of $X$. ⭐
• Give an example of a topological space that is not Hausdorff, but satisfies the $T_1$ axiom (i.e., all finite subsets are closed).
• Prove that if $X$ satisfies the $T_1$ axiom, then a point $x\in A\subset X$ is a limit point of $A$ if and only if every neighborhood of $x$ contains infinitely many points of $A$. (Hint: the reverse implication is immediate. For the forward implication, prove the contrapositive, and note that one can take the complement of $\{x\}$ in either $A$, $U$, or $A\cap U$ based on the definition of a limit point.)