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Definition

A topological space $X$ is said sequential when a subet $Y$ is not closed if and only if there exists a sequence $(x_n)_{n\in\Bbb N}$ in $Y$ convergint to a point $x_0$ of $X\setminus Y$.

So as exercise in the text Elementos de Topología General by Ángel Tamariz Mascarúa and Fidel Casarrubias Segura is asked to prove the following result.

Theorem 3.D.8.b

A function $f$ from a sequential space $X$ to a space $Y$ is continuous at $x_0$ if and only if every sequence $(x_n)_{n\in\Bbb N}$ converging to $x_0$ has immage $\big(f(x_n)\big)_{n\in\Bbb N}$ converging to $f(x_0)$.

So unfortunately I was not able to prove the previous theorem and moreover I did no find it in any text but in the text General Topology by Ryszard Engelking I found the following proposition

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which is surely similar to the theorem $3$.D.$8$.b but it is different. Perhaps can the theorem $3$.D.$8$.b be proved by the proposition 1.6.15? Moreover if this was true I point out that I did not understand the proof of $1.6.15$ because it is not clear how make the sequence in $f^{-1}[B]$ and then is not clear why $f^{-1}[B]$ is closed so that in this case I ask to explain these thigs. So could someone help me, please?

Antonio Maria Di Mauro
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  • Hey! See https://math.stackexchange.com/q/4380300/861687. – user264745 May 01 '22 at 14:26
  • @user264745 I see: however it seems that it does not help. Perhaps are you able to prove that a sequential space is metrizable? – Antonio Maria Di Mauro May 01 '22 at 14:31
  • @user264745 I do not think. – Antonio Maria Di Mauro May 01 '22 at 14:48
  • Rephrasing theorem 3.D.8.b: suppose $f$ be a function from a sequential space $X$ to a space $Y$. Show $f$ is continuous at $x_0$ if and only if every sequence $(x_n){n\in\Bbb N}$ converging to $x_0$ has immage $\big(f(x_n)\big){n\in\Bbb N}$ converging to $f(x_0)$. $\Rightarrow$ holds for general topological space. In this implication we don’t need sequential space condition. $\Leftarrow$ I’m also trying. But I can’t prove it. Can you show if $x\in \overline{A}$, then $\exists$ sequence in $A$ such that that sequence converge to $x$. – user264745 May 01 '22 at 15:56
  • @user264745 I already knew the first implication; the problem is exactly the second implication as you observed by your self. Anyway as here you can see I am not sure that for any point of $\ovelrine A$ there exists a sequence in $A$ converging there. – Antonio Maria Di Mauro May 01 '22 at 16:02
  • Can you offer bounty for this post? – user264745 May 03 '22 at 07:59
  • @user264745 I do not know: Oliver Díaz here effectively proved the theorem: the only one mistery if it is hols pointwise too but I think that this is genrally false because many text (e.g. Engelking) proves it only globally. Perhaps do you know if it holds pointwise too? – Antonio Maria Di Mauro May 03 '22 at 09:03
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    My knowledge of topology is not wide, you can see my posts(I’m beginner, essentially). Henno Brandsma is not active now. I will check/read Oliver Diaz answer. I can’t wait see how one would prove this problem. – user264745 May 03 '22 at 11:48
  • Oliver Diaz didn’t solve for continuity at a single point. I don’t think I have any chance. – user264745 May 05 '22 at 08:44
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    @user264745 I know that Oliver Diaz did not prove pointwise continuity and the same thing the professor Brandsma did here so surely pointwise continuity is not trivial provided obviously it does not fails. – Antonio Maria Di Mauro May 05 '22 at 08:47

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