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1. I changed the name of discrete space to discrete object such that it is now consistent with codiscrete object.

• CommentRowNumber2.
• CommentAuthorUrs
• CommentTimeFeb 29th 2012

Okay, good point.

• CommentRowNumber3.
• CommentAuthorMike Shulman
• CommentTimeMar 1st 2012

I added a hatnote.

• CommentRowNumber4.
• CommentAuthorTodd_Trimble
• CommentTimeMar 2nd 2016

In discrete object, I saw two mentions of the diagonal map “$X \times X \to X$”, so I made them both $X \to X \times X$.

The first paragraph under Discrete Geometric Spaces puzzled me, where it says, “the converse holds if $X$ satisfies the $T_0$ separation axiom” (i.e., if the diagonal map is open, then $X$ is discrete provided we assume $T_0$). I don’t understand why we need that assumption.

Suppose $X \to X \times X$ is an open map. In particular the image of the diagonal map is an open set in $X \times X$, i.e., for each $(x, x)$ there is a basic open $U \times V$ containing $(x, x)$ that is entirely contained in the diagonal. Thus the subset $\{x\} \times V$ of $U \times V$ would also be entirely contained in the diagonal, i.e., $(x, y) \in \{x\} \times V$ implies $x = y$, for any $y \in V$. So the open $V$ is the singleton $\{x\}$. (By similar reasoning, $U$ is also the singleton $\{x\}$.) So $\{x\}$ is open, for every $x \in X$. No separation axiom needed. Am I missing something?

• CommentRowNumber5.
• CommentAuthorZhen Lin
• CommentTimeMar 2nd 2016

A topological space has open diagonal if and only if it is discrete, indeed. I prefer this argument: for every $x \in X$, the intersection of the diagonal and $\{ x \} \times X$ is the singleton $\{ (x, x) \}$, hence $\{ x \}$ is open in $X$.

• CommentRowNumber6.
• CommentAuthorTodd_Trimble
• CommentTimeMar 2nd 2016

Oh, I see: the inverse image of the open $\Delta$ along $y \mapsto (x, y)$, for any given $x \in X$.

• CommentRowNumber7.
• CommentAuthorTodd_Trimble
• CommentTimeMar 2nd 2016

I went ahead and edited that point in.