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Kinda related: Set of branch points isn't discrete, but branch points are isolated?

Consider a closed and continuous map $f: X \to Y$ of any topological spaces.

Question: What are some conditions on $f, X$ and $Y$ such that $f$ maps discrete to discrete?

every image of a discrete space under a closed continuous map is discrete

and

every image of a discrete space under an open continuous map is also discrete

  • and also here (p. 2, in the proof of Theorem 17.6):

As $f$ is closed the image of a discrete set of points is discrete.

  • I guess neither answers the above question with 'any conditions'. I'm trying to understand which properties are of $f, X$ and $Y$ are relevant here.
BCLC
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    Add page number/line nr to the quotes? – Henno Brandsma Sep 05 '20 at 06:29
  • @HennoBrandsma Edited. thanks – BCLC Sep 05 '20 at 06:39
  • You don’t believe that the closed continuous image of a discrete space is discrete? Without conditions on $X$ or $Y$? – Henno Brandsma Sep 05 '20 at 08:10
  • @HennoBrandsma Ah well I didn't really bother to try it out since I wasn't sure if it was true. I'm not sure because (A) i couldn't find a reference that states outside the context of other spaces and because (B1) I mistakenly thought maps are open if and only if they are closed and of (B2) the other question...wait wait wait with my new edit on OPEN and continuous maps discrete to discrete, now I'm really not sure because of the question I linked just now – BCLC Sep 05 '20 at 08:13
  • @HennoBrandsma Actually, my question initially didn't include that open and continuous (ostensibly or otherwise under certain assumptions) maps discrete to discrete, but I happened to notice it when I was looking up the page numbers. – BCLC Sep 05 '20 at 08:15
  • Not all “folklore” facts get a reference. The same statement is trivial for open maps instead of closed. – Henno Brandsma Sep 05 '20 at 08:16
  • @HennoBrandsma Ok well now I know. Can you please prove it? Also, do you therefore disagree with Moishe Kohan in other question? Apparently Moishe Kohan seems to (implicitly) think that non-constant and holo does not necessarily map discrete to discrete, even though non-constant and holo implies open and continuous. – BCLC Sep 05 '20 at 08:24

1 Answers1

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Assume that $X$ is a discrete space.

If $f: X \to Y$ is open, then for every $y \in f[X]$ we write $y=f(x)$ and so $\{y\}= f[\{x\}]$ is open in $f[X]$ and so $f[X]$ is discrete. So for openness it's immediate.

If $f:X \to Y$ is closed and onto, then for $\{y\}$ in $Y$ we can say that

$$Y\setminus \{y\} = f[X\setminus f^{-1}[\{y\}]]$$ and hence $Y\setminus \{y\}$ is closed and $\{y\}$ is open in $Y$.

So we don't even need continuity.

Henno Brandsma
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  • wait what's up with these additional assumptions of $X$ discrete and $f$ surjective? I never said $X$ was discrete or that $f$ is surjective. I wanna know if for a discrete subset $A$ of $X$, we get that $f(A)$ is discrete in $Y$...i mean...just because $f: X \to Y$ is open doesn't mean either $f|_A: A \to Y$ or $g := \tilde{f|_A}: A \to f(A)$ is open right? – BCLC Sep 05 '20 at 09:12
  • @JohnSmithKyon you’re asking about images of discrete spaces, aren’t you? And we consider the image so wlog we can assume maps are onto as well. – Henno Brandsma Sep 05 '20 at 09:14
  • I'm asking about images of discrete subspaces actually....i mean, openness/closedness of $f$ implies openness/closedness of $f|_A$ or $g$? I know the answer is yes if we replace 'openness/closedness' with continuity – BCLC Sep 05 '20 at 09:17
  • @JohnSmithKyon I responded to the quoted statements from the papers you found. For open continuous maps the statement about discrete subspaces need not hold, of course. Trivial counterexamples exist. For closed maps, you probably need to strengthen it to proper or perfect maps, which might be enough, depending on the application. – Henno Brandsma Sep 05 '20 at 09:23
  • ok so maybe the 1st 2 quotes are not really relevant since they assume the whole of the domain is discrete. but what about the 3rd quote? (note: the 2nd quote was the one i added later on. the 3rd quote, formerly the 2nd quote, was there at the start) i believe the 3rd quote does not assume the whole of the domain is discrete. in this case, what are the conditions on $f, X$ and $Y$ that ensure image of discrete subspace is discrete? – BCLC Sep 05 '20 at 09:29
  • @JohnSmithKyon There they mean ( I think) by a "discrete set of points" a closed set such that it has the discrete topology as a subspace, and then the closed argument from the answer applies, as the two-sided restriction is then also a closed map. – Henno Brandsma Sep 05 '20 at 09:33
  • Henno Brandsma, thanks for the replies thus far (1) 2-sided refers to the $g$? (2) what about 'Now, the ramification and branch points must form a discrete set' in here or Federico Fallucca's answer here please? – BCLC Sep 05 '20 at 09:51