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I've read the definition of a topological Hausdorff space (two distinct points have disjoints neighbourhoods) and of a disconnected space (it is the union of two disjoint nonempty open sets)

Now I think an equivalence definition of a topological space being Hausdorff would be this:

Prop 1) A topological space $X$ is Hausdorff iff every two point subset $\{x,y\} \subseteq X$ (with $x \neq y$) is a disconnected subspace of $X$.

Proof. $\Rightarrow$) $X$ is Hausdorff. If $X = \emptyset$ or a singleton it is valid. Else, for any $\{x,y\} \subseteq X$ with $x\neq y$ we take the disjoints (open) neighbourhoods $x \in E$, $y \in V$, and so by the subspace topology $\{x\}$ and $\{y\}$ are open sets of $\{x,y\}$, so $\{ \{x\}, \{y\} \}$ is a disconnection, and $\{x,y\}$ is disconnected.

$\Leftarrow$) If every $\{x,y\} \subseteq X$ with $x\neq y$ is disconnected, then there are (open) neighbourhoods $E,V$ such that $x \in E$, $y \in V$ and $E \cap V = \emptyset$. This are disjoint neighbourhoods and is valid $\forall x,y \in X$, so $X$ is Hausdorff.

Is this ok? If it is, it's strange to me that it isn't usually mentioned in the wikipedia/textbooks (or I haven't seen it at least).


Edit: It seems Proposition 1 is false. I'll try to show the following

Prop 2) A topological space $X$ is $T_1$ iff every two point subset $\{x,y\} \subseteq X$ (with $x \neq y$) is a disconnected subspace of $X$.

Proof. $\Rightarrow$) $X$ is $T_1$ so for every $x,y \in X$, $x$ has a neighbourhood $E$ not containing $y$, and $y$ has a neighbourhood $V$ not containing $x$. So $\{x\}$ and $\{y\}$ are open in $\{x,y\}$, non empty and disjoint, so $\{x,y\}$ are disconnected.

$\Leftarrow$) Every $\{x,y\}$ is disconnected, let $E,V$ be a disconnection, then there exists open sets $E',V'$ of $X$ such that $E = E' \cap \{x,y\} = \{x\}$ and $V = V' \cap \{x,y\} = \{y\}$. This $E', V'$ are neighbourhood of each point not containing the other, and is valid for every $x, y \in X$, so $X$ is $T_1$.

Is this ok now?

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  • $\begingroup$ it looks like it is ok ank many people use this king of linkage between T2 and the the disconnection idea. $\endgroup$
    – Adelafif
    Jul 11, 2015 at 9:18

1 Answer 1

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This isn't quite right. The problem is that in the reverse direction, you don't know that $E$ and $V$ are open in $X$; you only know that they are open in the subset topology on $\{x,y\}$. This means that there are open sets $E',V'\subseteq X$ such that $E=E'\cap \{x,y\}$ and $V=V'\cap\{x,y\}$. But now you don't know that $E'$ and $V'$ are disjoint, because they might intersect at points of $X$ other than $x$ and $y$.

In fact, your condition is equivalent to the $T_1$ axiom, not the Hausdorff axiom. Can you prove this?

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    $\begingroup$ I think I understand what you mean. For example if $X = \{1,2,3\}$ and the topology is $T = \{ \{1,2,3\}, \{1,2\}, \{1,3\}, \{1\}, \emptyset \}$ then $\{2,3\} \subseteq X$ is a disconnected subspace, but the space is not Hausdorff because there are not disjoint neighbourhoods for $2$ and $3$ $\endgroup$
    – dami
    Jul 11, 2015 at 9:30
  • $\begingroup$ @dami Exactly. Another example is if $X$ is the real numbers, and we take the topology where the closed sets are all finite subsets (equivalently, where the open sets are all cofinite subsets; i.e., subsets whose complement is finite). $\endgroup$ Jul 11, 2015 at 9:54

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