For first countable topological spaces, it is enough to consider convergence of sequences to determine closure of sets and continuity of functions. The following result can be found in several Topology books (Kelley's Genral Topology for instance. The section on Topology of "Hitchhiker's guide to Infinite Dimensional Analysis" by the late Aliprantis (a fantastic read) covers this in a very elegant way)
Theorem: If $(X,\tau)$ is first countable, then:
- $X$ is Hausdorff iff any convergent sequence in $X$ has a unique limit.
- A point $x\in X$ is a cluster point of a sequence $\{x_n:n\in\mathbb{Z}_+\}$ iff there exists a subsequence that converges to $x$.
- A sequence $x_n$ converges to $x$ iff every subsequence converges to $x$.
- $x\in\overline{A}$ iff there is a sequence $x_n\in A$ that converges to $x$.
- For any topological space $(Y,\tau')$ and function $f:X\rightarrow Y$, $f$ is continuous at $x$ iff for any sequence $x_n\xrightarrow{n\rightarrow\infty} x$, $f(x_n)\xrightarrow{n\rightarrow\infty} f(x)$.
- More generally, for any topological space $(Y,\tau')$ and function $f:X\rightarrow Y$, $\lim_{u\rightarrow x}f(x)=L$ iff $\lim_{n\rightarrow\infty}f(x_n)=L$ for any sequence $\{x_n:n\in\mathbb{N}\}\subset X$ such that $\lim_nx_n=x$.
Here is a sketht of the proof
By hypothesis, any point $x\in X$ has a countable local base $\mathscr{V}_x=\{V_n:n\in\mathbb{N}\}$ and, by setting $U_n=\bigcap^n_{j=1} V_j$ if necessary, we may assume that $V_n\subset V_{n+1}$ for all $n\in\mathbb{N}$.
(1) Since any sequence is a net, only sufficiency remains to be proved. Suppose any convergent sequence in $X$ has a unique limit. Let $x$ and $y$ be points in $X$ and let $\{V_n:n\in\mathbb{N}\}$ and $\{U_n:n\in\mathbb{N}\}$ be decreasing local neighborhoods of $x$ and $y$ respectively. If $V_n\cap U_n\neq\emptyset$ for all $n\in\mathbb{N}$ then we can choose $x_n\in V_n\cap U_n$. The sequence $\{x_n:n\in\mathbb{N}\}$ converges to both $x$ and $y$. Therefore, $x=y$.
(2) Since a subsequence of a sequence is a subnet of the sequence, only necessity remains to be proved. Suppose $x$ is a cluster point of the sequence $\{x_n:n\in\mathbb{N}\}$. There is $n_1\geq 1$ such that $x_{n_1}\in V_1\in \mathscr{V}_x$. Having found $x_{n_1},\ldots, x_{n_k}$ such that $n_1<\ldots < n_k$ and $x_{n_j}\in V_j$ we choose $x_{n_{k+1}}\in V_{k+1}$ such that $n_{k+1}\geq n_k+1$, which is possible since $x$ is a cluster point of $\{x_n:n\in \mathbb{N}\}$. Therefore, $\{x_{n_k}:k\in\mathbb{N}\}$ is a subsequence that converges to $x$.
(3) This statement is trivial, try to complete it.
(4) Since any sequence is a net, only necessity remains to be proved.
If $x\in \overline{A}$ then $V_n\cap A\neq\emptyset$ for each $V_n\in\mathscr{V}_x$. Choosing $x_n\in V_n\cap A$ for each $n\in\mathbb{N}$, we obtain a sequence $x_n\xrightarrow{n\rightarrow\infty} x$.
(5) Since any sequence is a net, only sufficiency remains to be proved. Suppose $f(x_n)\xrightarrow{n\rightarrow\infty} f(x)$ whenever $x_n$ is a sequence with $x_n\xrightarrow{n\rightarrow\infty} x$. If $f$ fails to be continuous at $x$, then there is a neighborhood $U\in\mathcal{V}_{f(x)}$ such that for any $n\in\mathbb{N}$ there is $x_n\in V_n$, $V_n\in\mathscr{V}_x$, with $f(x_n)\notin U$. Then $x_n$ is a sequence converging to $x$ for which $f(x_n)\nrightarrow f(x)$. This is a contradiction.
(6) By replacing $f(x)$ by $L$ in the proof of (5), the remaining of that roof carries over.
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.
Best Answer
The problem is that if a space isn’t first countable, its topology may not be completely determined by its convergent sequences. Here’s a relatively straightforward example.
Let $\tau$ be the order topology on $\omega_1+1$, and let
$$\tau'=\big\{\{\alpha\}:\alpha\in\omega_1\big\}\cup\big\{\{\omega_1\}\cup(\omega_1\setminus C):C\subseteq\omega_1\text{ and }|C|\le\omega\big\}\cup\{\varnothing\}\;;$$
$\tau'$ is a topology on $\omega_1$ finer than $\tau$. (In other words, we just isolate each $\alpha\in\omega_1$ and let the point $\omega_1$ keep its usual nbhds.) Let $X$ be $\langle\omega_1+1,\tau'\rangle$ and $Y$ be $\langle\omega_1+1,\tau\rangle$.
Note that the only convergent sequences in $X$ are the ones that are eventually constant, so any function from $X$ to $Y$ takes convergent sequences in $X$ to convergent sequences in $Y$; it will not be surprising, then, that some of those functions are not continuous. Now let
$$f:X\to Y:\alpha\mapsto\begin{cases} 0,&\text{if }\alpha<\omega_1\\ \omega_1,&\text{otherwise;} \end{cases}$$
then $Y\setminus\{0\}$ is open in $Y$, but its inverse image under $f$ is $\{\omega_1\}$, which is not open in $X$, so $f$ is not continuous.