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1. • CommentRowNumber1.
   • CommentAuthorStephan A Spahn
   • CommentTimeFeb 29th 2012
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   Author: Stephan A Spahn
   Format: MarkdownItexI changed the name of [[discrete space]] to [[discrete object]] such that it is now consistent with [[codiscrete object]].

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

2. • CommentRowNumber2.
   • CommentAuthorUrs
   • CommentTimeFeb 29th 2012
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   Author: Urs
   Format: MarkdownItexOkay, good point.

   Okay, good point.

3. • CommentRowNumber3.
   • CommentAuthorMike Shulman
   • CommentTimeMar 1st 2012
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   Author: Mike Shulman
   Format: MarkdownItexI added a hatnote.

   I added a hatnote.

4. • CommentRowNumber4.
   • CommentAuthorTodd_Trimble
   • CommentTimeMar 2nd 2016
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   Author: Todd_Trimble
   Format: MarkdownItexIn [[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?

   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?

5. • CommentRowNumber5.
   • CommentAuthorZhen Lin
   • CommentTimeMar 2nd 2016
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   Author: Zhen Lin
   Format: MarkdownItexA 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 $X$.

   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 $X$.

6. • CommentRowNumber6.
   • CommentAuthorTodd_Trimble
   • CommentTimeMar 2nd 2016
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   Author: Todd_Trimble
   Format: MarkdownItexOh, I see: the inverse image of the open $\Delta$ along $y \mapsto (x, y)$, for any given $x \in X$.

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

7. • CommentRowNumber7.
   • CommentAuthorTodd_Trimble
   • CommentTimeMar 2nd 2016
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   Author: Todd_Trimble
   Format: MarkdownItexI went ahead and edited that point in.

   I went ahead and edited that point in.