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1. • CommentRowNumber1.
   • CommentAuthorStephan A Spahn
   • CommentTimeFeb 29th 2012
   • PermaLink
   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 “”, so I made them both .

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

   Suppose is an open map. In particular the image of the diagonal map is an open set in , i.e., for each there is a basic open containing that is entirely contained in the diagonal. Thus the subset of would also be entirely contained in the diagonal, i.e., implies , for any . So the open is the singleton . (By similar reasoning, is also the singleton .) So is open, for every . 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 , the intersection of the diagonal and is the singleton , hence is open in .

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 along , for any given .

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.