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Consider topological spaces $(S,\,{\cal{T}})\,$, $(S^{\,\prime},\,{\cal{T}}^{\,\prime})\;$ and a continuous map $F: S\longrightarrow S^{\,\prime}$.

The limit of a sequence $\{x_n\}$ is a point $x\in S$, every neighbourhood of which contains all the points of the sequence, except at most finitely many of them.

  1. My understanding is that if $\{F(x_n)\}$ is convergent then $\{x_n\}$ is convergent, provided $F$ is continuous.
    Please correct me if I am wrong.
  2. Would the reverse statement be correct, that a continuous map sends a convergent sequence to a convergent one? I suspect, it won't, unless $F$ is an open function, i.e., maps open sets to open sets. But I am not sure...
Michael_1812
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    Continuity always implies sequential continuity but not the converse. – SoG Sep 16 '23 at 16:56
  • Souav Ghosh By sequentil continuity, do you imply that the preimage converges? Also, if you could shape your comment into an answer, that would give me an opportinity to accept it, so you get a full credit in points. – Michael_1812 Sep 16 '23 at 16:59
  • $x_n\to x\implies f(x_n) \to f(x) $ – SoG Sep 16 '23 at 17:00
  • @SouravGhosh How to prove this? Is a reverse statement correct also? – Michael_1812 Sep 16 '23 at 17:02
  • Suppose $F:S\to S'$ continuous and $s_n\to s $ in $(S, \tau) $ . Then show that $F(s_n) \to F(s) $ in $(S', \tau') $ – SoG Sep 16 '23 at 17:05
  • @SouravGhosh This is exactly the subject of my question: how to prove that $F(x_n) \longrightarrow F(x)$ ? – Michael_1812 Sep 16 '23 at 17:06
  • For converse, take $\textrm{Id}:(\Bbb{R}, \tau_{\textrm{cocountable}}) \to(\Bbb{R}, \tau_{\textrm{euclidean}}) $ – SoG Sep 16 '23 at 17:08
  • @SouravGhosh So your answer to my 2nd question is negative, correct? How about my 1st question? Not being a mathematician, I don't know what cocountable is. A more detailed answer understandable to a layman would be most appreciated. – Michael_1812 Sep 16 '23 at 17:09
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    See here and here. – SoG Sep 16 '23 at 17:12
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    "the preimage ${x_n}$ of a converging sequence ${F(x_n)}$" makes no sense, unless $F$ is invertible. But I suppose what you mean to ask in part (i) is, if $(F(x_n))$ converges, does $(x_n)$ necessarily converge? – TonyK Sep 16 '23 at 17:14
  • @TonyK Yes, this is what I wanted to ask. Corrected my question. Thank you. – Michael_1812 Sep 16 '23 at 17:16
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    A constant map is always continuous, and sends every sequence to a (trivial) convergent sequence. – Steven Clontz Sep 16 '23 at 17:47

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