Theorem 30.1 (b) in Munkres' TOPOLOGY, 2nd ed: The sequential criterion for continuity

Question

Here is Theorem 30.1 in the book Topology by James R. Munkres, 2nd edition:

Let $X$ be a topological space.

(a) Let $A$ be a subset of $X$. If there is a sequence of points of $A$ converging to $x$, then $x \in \overline{A}$; the converse holds if $X$ is first-countable.

(b) Let $f \colon X \to Y$. If $f$ is continuous, then for every convergent sequence $x_n \to x$ in $X$, the sequence $f\left(x_n\right)$ converges to $f(x)$. The converse holds if $X$ is first-countable.

And, here is the definition of a topological space to be first-countable.

A space $X$ is said to have a countable basis at $x$ if there is a countable collection $\mathscr{B}$ of neighborhoods of $x$ such that each neighborhood of $x$ contains at least one of the elements of $\mathscr{B}$. A space that has a countable basis at each of its points is said to satisfy the first countability axiom, or to be first-countable.

Finally, here is Theorem 18.1 in Munkres:

Let $X$ and $Y$ be topological spaces; let $f \colon X \longrightarrow Y$. Then the following are equivalent:

(a) $f$ is continuous.

(b) For every subset $A$ of $X$, one has $f\left(\overline{A}\right) \subset \overline{f(A)}$.

(c) For every closed set $B$ of $Y$, the set $f^{-1}(B)$ is closed in $X$.

(d) For each $x \in X$ and each neighborhood $V$ of $f(x)$, there is a neighborhood $U$ of $x$ such that $f(U) \subset V$.

Thus we have the following result:

Let $X$ and $Y$ be topological spaces, and let $f \colon X \longrightarrow Y$ be a function. Then $f$ is continuous if and only if, for every subset $A$ of $X$, we have $$ f\left( \overline{A} \right) \subset \overline{ f(A)}. $$

Here is the definition of continuity of a function at a point of its domain.

As my understanding goes, using this result, together with Theorem 30.1 (a), we can prove the converse of Theorem 30.1 (b). Am I right?

Here is the definition of continuity of a mapping between topological spaces at a point of its domain:

Let $X$ and $Y$ be topological spaces, let $f \colon X \longrightarrow Y$ be a function, and let $p$ be a point of $X$. Then $f$ is said to be continuous at $p$ if, for every open set $V$ in $Y$ such that $f(p) \in V$, there exists an open set $U$ in $X$ such that $$p \in U \subset f^{-1}(V). $$

Now my question is, can we also prove the following?

Let $X$ and $Y$ be topological spaces, let $f \colon X \longrightarrow Y$ be a function, and let $p$ be a point of $X$. Suppose $X$ has a countable (local) basis at $p$; suppose also that, for every sequence $\left( p_n \right)_{n \in \mathbb{N}}$ of points of $X$ converging to point $p$ in $X$, the image sequence $\left( \, f\left(p_n\right) \, \right)_{n \in \mathbb{N}}$ converges to $f(p)$ in $Y$. Then $f$ is continuous at point $p$.

My Attempt:

Let $V$ be an open set in $Y$ such that $f(p) \in V$. Then . . .

What next? How to proceed from here?

In short, my question is, can we also state Theorem 30.1 (b) --- especially the converse part thereof --- "on a local scale"? Or, is this a result of a purely global nature? If so, any counter-examples, please!

PS. We can prove this final result by contradiction as follows:

Suppose that our function $f \colon X \longrightarrow Y$ is not continuous at point $p \in X$. Then, by definition, there exists an open set $V_0$ of $Y$ containing $f(p)$ such that, for every open set $U$ of $X$ containing $p$, we have $$ f(U) \not\subset V_0, $$ or in other words, $$ U \not\subset f^{-1} \left( V_0 \right), $$ and hence $$ U \setminus f^{-1} \left( V_0 \right) \neq \emptyset. \tag{1} $$

Now let $\left\{ B_n \colon n \in \mathbb{N} \right\}$ be a countable local basis at point $p$, and for each $n \in \mathbb{N}$ let us put $$ U_n \colon= B_1 \cap \cdots \cap B_n. \tag{2} $$ Then each $U_n$ is an open set of $X$ containing $p$, and using (1) above we of course have $$ U_n \setminus f^{-1} \left( V_0 \right) \neq \emptyset, $$ so that there exists a point $p_n \in U_n \setminus f^{-1} \left( V_0 \right)$.

From (2) above we also have $$ U_1 \supset U_2 \supset \cdots. \tag{3} $$ We shw that the sequence $\left( p_n \right)_{n \in \mathbb{N}}$ obtained in the preceding paragraph convereges to $p$ in $X$. Let $U$ be any open set of $X$ containing $p$. Since the collection $\left\{ B_n \colon n \in \mathbb{N} \right\}$ is a countable local basis at $p$, therefore there exists a natural number $N_p$ such that $$ p \in B_{N_p} \subset U. \tag{4} $$ So we can conclude that, for every natural number $n > N_p$ we have $$ p_n \in U_n \subset U_{N_p} \subset B_{N_p} \subset U, $$ and hence $p_n \in U$ for every $n \geq N_p$, from which it follows that the sequence $\left( p_n \right)_{n \in \mathbb{N}}$ converges in $X$ to the point $p$.

However, for each $n \in \mathbb{N}$, since $p_n \not\in f^{-1} \left( V_0 \right)$, therefore $f \left( p_n \right) \not\in V_0$, and since $V_0$ is an open set of $Y$ containing $f(p)$, we can conclude that the image sequence $\left( \, f \left( p_n \right) \, \right)_{n \in \mathbb{N} }$ cannot converge in $Y$ to the point $f(p)$.

This completes the proof.

Am I right?

Is this proof correct and clear enough in each and every part thereof? Or, are there any issues in it?

Answer 1

So called "sequential continuity" does not characterize continuity. As the theorem says in Munkres, you only get the converse when $X$ is first countable.

For a counterexample, see the following

Sequentially continuous but not continuous

You can characterize continuity using nets instead of sequences though. An explanation of this is also in Munkres, although it is relegated to the exercises.

The other statement you want to prove is true though.

Let $f:X\rightarrow Y$ be a function and $p\in X$ a point admitting a countable neighbourhood basis $\{U_{n}\}_{n\in\mathbb{N}}$. Then $f$ is continuous at $p$ if and only if for every sequence $(x_{n})$ in $X$ that converges to $p$ we have that $(f(x_{n}))$ converges to $f(p)$.

We may assume that the neighbourhood basis of $p$ is such that $n<m$ implies that $U_{n}\subseteq U_{m}$.

Assume that $f$ satisfies the sequential condition at $p$. We then let $V\subseteq Y$ be an open neighbourhood of $f(p)$. Assume towards a contradiction that for all $n$ we have that $f(U_{n})\not\subseteq V$. Then for each $n\in\mathbb{N}$ we may choose an $x_{n}\in U_{n}$ such that $f(x_{n})\notin V$. However, the sequence $(x_{n})$ converges to $p$ (easily checked). Then $(f(x_{n}))$ must converge to $f(p)$, but as $f(x_{n})\notin V$ for all $n$, this is impossible. This is a contradiction. Therefore, there is some $n\in\mathbb{N}$ for which $f(U_{n})\subseteq V$, establishing continuity at $p$.

Now assume that $f$ is continuous at $p$ in the sense that for all open $V\subseteq Y$ containing $f(p)$ there is some open $U\subseteq X$ such that $p\in U$ and $f(U)\subseteq V$. We then let $(x_{n})$ be a sequence in $X$ converging to $p$. We need to show that $(f(x_{n}))$ converges to $f(p)$. If$(f(x_{n}))$ does not converge to $f(p)$ there is an open neighbourhood $V\subseteq Y$ of $f(p)$ that doesn't contain infinitely many of the points $f(x_{n})$. By assumption there is an open $U\subseteq X$ containing $p$ such that $f(U)\subseteq V$. Because $(x_{n})$ converges to $p$ we have that there is an $N\in\mathbb{N}$ such that for all $m\geq N$ we have that $x_{m}\in U$. That is, there are only finitely many $x_{n}$ that are not in $U$. Because $f(U)\subseteq V$ we have that there are only finitely many $f(x_{n})$ not in $V$, contradicting there being infinitely many $f(x_{n})$ not in $V$. Therefore we must have that $(f(x_{n}))$ converges to $f(p)$.

Edit: I'm aware that the image of the sequence $(x_{n})$ may be finite. When I say "$V$ doesn't contain infinitely many of the points $f(x_{n})$" I mean that there is an infinite subset $K\subseteq\mathbb{N}$ such that $f(x_{s})\notin V$ for all $s\in K$.