# General Topology – Sequence Convergence in Sequential Spaces

examples-counterexamplesgeneral-topologysolution-verification

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.

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 $$Y$$.

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?

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.

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.

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.