Prove or disprove that $x_0\in\operatorname{cl} Y$ iff there exists a sequence $x_n$ in $Y\subseteq X$ converging to $x_0$ when $X$ is sequential.

Question

The following is an exercise from Elementos de Topología General by Ángel Tamariz Mascarúa and Fidel Casarrubias Segura.

First a couple of definitions:

Definiton A topological space $X$ is Frechét-Urysohn if for very $A\subset X$ and $x\in\overline{A}$, there is a sequence $(x_n:n\in\mathbb{N})\subset A$ that converges to $x$.

Definition A topological space $X$ is said to be sequential if for any $Y\subset X$, $Y$ is not closed if and only if there exists $x\in X\setminus Y$ and a sequence $(x_n: n\in\Bbb N)\subset Y$ with converges to $x$.

Problem 7 and 8 in that textbook say:

• 7: Check that a first countable topological space is Frechét-Uryshohn, and that this property is inherited by any subspace of $X$.

• 8(a) Show that every closed subset of a sequential space is also sequential, and that every Fréchet-Uryshohn space is sequential.

• 8(b) Check that propositions 3.41 and 3.42 (below) hold if the assumption first countable is replaced by sequential.

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Here are the statement of the Propositions mentioned in the problem:

Proposition 3.41: If $X$ is first countable and $E\subset X$, then $x\in \overline{E}$ (closure of $E$) iff there is a sequence $(x_n:n\in\mathbb{N})\subset E$ that converges to $x$.

Proposition 3.42: Let $f:X\rightarrow Y$ be a map between topological spaces. If $X$ is first countable and $x\in X$, then $f$ is continuous at $x$ if and only if for every sequence $(x_n:n\in\mathbb{N})$ with $x_n\xrightarrow{n\rightarrow\infty}x$, $f(x_n)\xrightarrow{n\rightarrow\infty}y$ in $Y$.

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So I am really confused about the preceding theorem becasue if it is was true then it seem to me it would prove that a sequential space is a Fréchet-Uryshon space but as here showed this is false. So first of all to follow I put a direct reference of the text I mentioned hoping I understood bad Spanish text.

Anyway I tried to prove the statement and surprisingly it seems true. So to follow my proof attempt.

proof $\,8$.b.$1.\,\,$

So if $(x_n)_{n\in\Bbb N}$ is a sequence in $Y$ converging to $x_0$ then by convergence definition for any neighborhood $V$ of $x_0$ there exist $n_V\in\Bbb N$ such that $$ x_n\in V $$ for all $n\ge n_V$ but if $x_n$ is in $Y$ for all $n\in\Bbb N$ then this means that $V\cap Y$ is not empty, that is $x_0$ is adherent to $Y$.

Conversely let be $x_0\in\operatorname{cl}Y$ and we let use sequentiality to make a sequence on $Y$ converging to $x_0$. So if $x_0$ is an isolated point of $Y$ then trivially the position $$ x_n:=x $$ for all $n\in\Bbb N$ defines sequence in $Y$ converging to $x_0$ so we suppose that $x_0$ is an accumulation point for $Y$. Now if $x_0$ is an accumulation point for $Y$ then as here showed the identity $$ \operatorname{cl}\big(\operatorname{cl}Y\setminus\{x_0\}\big)=\operatorname{cl} Y $$ holds and thus we conclude that $\operatorname{cl}Y\setminus\{x_0\}$ is not closed so that by sequentiality there exists a sequence $(x_n)_{n\in\Bbb N}$ in $\operatorname{cl}Y\setminus\{x_0\}$ converging to $x\notin \operatorname{cl}Y\setminus\{x_0\}$. So if $x_n$ is in $\operatorname{cl}Y\setminus\{x_0\}$ for all $n\in\Bbb N$ then by the first implication $x$ must be in $\operatorname{cl}\big(\operatorname{cl}Y\setminus\{x_0\}\big)$ that is in $\operatorname{cl} Y$ and so $x$ must be equal to $x_0$ because the unique element of $\operatorname{cl} Y$ not in $\operatorname{cl}Y\setminus\{x_0\}$ is $x_0$.

So is the proposition $8$.b.$1$ true? if it is true then is the proof I gave correct? Could someone help me, please?

Answer 1

I write this to summarize my previous comments and Brian Scott's insightful comments.

Indeed, there must be a typo in the textbook to which you are referring. You may want to contact one of the authors to see if they keep tabs on errata. What I think problem 8 should have ask is to check whether propositions 3.41 and 3.42 hold when first countable is replaced by sequential.

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Suppose $(X,\tau)$ is has the Frechét-Urysohn property. If $A\subset X$ is not closed, then there exists $a\in \overline{A}\setminus A$. Consequently there exists a sequence $(x_n:n\in\mathbb{N})\subset A$ such that $x_n\xrightarrow{n\rightarrow\infty}a$. This implies that $(X,\tau)$ is sequential.

Prop 3.41 implies that every first countable space has the Frechét-Urysohn property.

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8b:

• Regarding Prop. 3.41. As you pointed out, proposition 3.41 does not hold for general sequential spaces. Otherwise, every sequential space would have the Frechét-Urysohn-Urysohn property; however, there are Frechét-Urysohn-Urysohn spaces that are not sequential (the Arens space)
• Regarding Prop 3.42. Suppose $(X,\tau)$ is sequential and $f:X\rightarrow Y$ is a map such that for any $x\in X$ and sequence $x_n\xrightarrow{n\rightarrow\infty}a$ implies $f(x_n)\xrightarrow{n\rightarrow\infty}f(x)$. If $f$ is not continuous, then there is a closed set $F\subset Y$ such that $f^{-1}(F)$ is not closed in $X$. Then, there is $x\in X\setminus f^{-1}(F)=f^{-1}(Y\setminus F)$ and a sequence $(x_n:n\in\mathbb{N})\subset f^{-1}(F)$ such that $x_n\rightarrow x$. Then, $(f(x_n):n\in\mathbb{N})\subset F$, and by assumption, $f(x_n)\xrightarrow{n\rightarrow\infty}f(x)$. Then $f(x)\in F$ contradicting the fact that $x\notin f^{-1}(F)$. This shows that Proposition 3.42 holds if first countable is replaced by sequential.