In topological spaces, which condition is necessary for a sequentially continous function $f: (X,\tau_x) \rightarrow (Y,\tau_y)$ to be continous?
I have tried to prove this making the space X be $T_1$ and then making it Hausdorff but I don't get the answer. For example making $\tau_x$ the topology of the complements of countable sets is $T_1$ and $f(x) = x$ is sequentially continuous but not continuous taking $X=Y= \mathbb{R}$ and $\tau_y$ being the usual topology in $\mathbb{R}.$
A function $f$ between topological spaces is sequentially continuous if the image of every convergent sequence is a sequence which converges to the image of the limit.
Continuity always implies sequential continuity: Suppose $x_n\rightarrow x$. Then if $U$ is any open neighborhood of $f(x)$, $f^{-1}(U)$ is a neighborhood of $x$ which (by continuity of $f$) is open. Because $x_n\rightarrow x$, every neighborhood of $x$ contains a tail of the sequence. In particular, $f^{-1}(U)$ contains a tail of the sequence. Hence $f(f^{-1}(U))\subseteq U$ contains the image of the tail, which is a tail of the image. Since $U$ was arbitrary, we have that every neighborhood $U \ni f(x)$ contains a tail of the image of the sequence. This means that $f(x_n)\rightarrow f(x)$.
A space is called first countable if for each point $x$, there is a countable collection of open sets around $x$ such that any set around $x$ contains some member of the collection.
In a first-countable space, the entire topology (open and closed sets) can be characterized in terms of sequences. Moreover, for functions whose domain is a first-countable space, sequential continuity implies continuity: suppose $f$ is sequentially continuous. Pick any point $x$; we will show that $f$ is continuous at $x$. Suppose $x_n\rightarrow x$. Then $f(x_n)\rightarrow f(x)$ by sequential continuity. Hence every neighborhood $U\ni f(x)$ contains a tail of the image of the sequence. Hence $f^{-1}(U)$ contains a tail of the sequence itself. We can't assume that $f^{-1}(U)$ is open (because we haven't proved that $f$ is continuous), but by first countability, we know that $f^{-1}(U)$ contains an open set $V \ni x$. We know that $V\subseteq f^{-1}(U)$, so $f(V)\subseteq U$, so $f$ is continuous at $x$.
More generally, we use the term sequential spaces to refer to the collection of topological spaces for which sequential continuity implies continuity. All first countable spaces are sequential spaces, as we have shown. And there are others: consider $\mathbb{R}/\mathbb{Z}$; the real line with its usual topology, but where the integers have been unified to a single point. You can show that this space is not first countable, but it is sequential nonetheless.
$\mathbb{R}/\mathbb{Z}$ isn't first countable: you can visualize the space as a countable number of circular loops splayed out in 3D space; the circles all meet at a single origin point 0. Each loop corresponds to an interval between integers $[n,n+1]$. The space fails to be first countable because we can't find the required countable basis for the origin point 0. For contradiction, suppose we have a collection $C$ of open sets that we want to prove is a countable basis for 0. Consider the loops of this space: how many members of $C$ does each loop contain? If $C$ is countable, there must be at least one loop $[n,n+1]$ that contains finitely many members of $C$. But then we can find a set $[n+\frac{1}{2} - \epsilon, n+\frac{1}{2} + \epsilon]$ which is smaller than the smallest such member of $C$. This set is a neighborhood of 0 but contains no member of $C$, so $C$ is not a basis for the point 0.
I don't know if there is any other surprising defining characteristic of sequential spaces besides its straightforward definition.
We have to consider sequential domain spaces $X$.
A space $X$ is sequential when for all sequentially closed subsets $A$ of $X$, $A$ is closed in $X$.
$A \subseteq X$ is said to be sequentially closed, iff for all sequences $(a_n)$ in $A$ (i.e. all $a_n \in A$) such that $a_n \to x$ (in $X$), we have $x \in A$ as well. Note that always all closed sets are sequentially closed. But e.g. in the co-countable topology all convergent sequences are eventually constant, which implies all subsets of $X$ are sequentially closed (but not all subsets are closed).
It is well-known that all first countable spaces are sequential (this includes all metric spaces).
Theorem: if $f:X \to Y$ is sequentially continuous, and $X$ is sequential then $f$ is continuous.
Proof: let $C$ be closed in $Y$, we'll show that $A = f^{-1}[C]$ is closed in $X$. For this we only need to show it is sequentially closed. So let $a_n \in A$ be a sequence such that $a_n \to x$. Then $f(a_n ) \to f(x)$ by sequential continuity. But $f(a_n) \in C$ by definition of $A$, so as $C$ is closed, $f(x) \in C$, which says $x \in f^{-1}[C] = A$, as required. So $f^{-1}[C]$ is closed in $X$ for all closed $C$ in $Y$, hence $f$ is continuous.
Suppose that $X$ obeys the conclusion of the theorem (all sequentially continuous maps on $X$ are continuous). If $X$ were not sequential we'd have a subset $A$ of $X$ that is sequentially closed but not closed. Consider $\tau'$: the topology generated by $\tau_X$ and $X\setminus A$. I think we can show that the identity $(X, \tau_X)$ to $(X,\tau')$ is sequentially continuous, and it is certainly not continuous.
So in a way, being a sequential space is the natural notion here to consider.
Proposition: if $(X, \mathscr T)$ and $(Y, \mathscr S)$ are topological spaces, $X$ is first countable, and $f: X → Y$ is sequentially continuous, then $f$ is continuous.
Proof: by contra-positive
Recall,
A: A (countable) local base for a topological space $(X, \mathscr T)$ at a point $x$ is a (countable) collection of open sets $\{X_i \in \mathscr T\}$ containing $x$, such that any open set $O$ that contains $x$ contains one of the sets $X_i$.
B: if there is a countable local base at $x$, then there is also a nested (descending sequence of sets) countable local base.
C: A topological space is first countable if for every point $x \in X$ there is a (possibly different) (nested) countable local base.
D: if $(X, \mathscr T)$ and $(Y, \mathscr S)$ are topological spaces, a function $f: X → Y$ is discontinuous at a point $x \in X$ if for some open set $V$ with $f(x) \in V$ then for every open set $U \in X$ with $x \in U$, $U \not\subset f^{-1}(V)$: equivalently $f (U) ⊄ V$.
E: An (infinite) sequence of points $(x_n) \in X$ converges to a limit $x \in X$ if for every open set $O$ containing $x$ there is $N$ such that for all $n > N$ then $x_n \in O$.
Suppose that $f: X → Y$ is not continuous $\implies$ it is not continuous at some point $x$.
Then by (D) there is some open set $V ⊂ Y$ with $f(x) \in V$ and for every open set $U \subset X$ with $x \in U, f (U) \not \subset V$.
Since $X$ is first countable, there is a nested countable local base $(X_i)$ at $x$, each $X_i$ is open, and by the previous sentence $f (X_i) \not \subset V$.
So for each of the nested $X_i, f (X_i) \not \subset V \implies$ there is some $x_i \in X_i$ with $f(x_i) \not \in V$.
Pick one from each $X_i$ and consider the sequence $(x_i)$.
By (E), the sequence $(x_i)$ converges to x since....
For any open set $O \in X$ which contains $x$ there is some $X_i$ with $x \in X_i \subset O$ (definition of a local base)
And for $j \ge i$ then because the local base is nested, all $x_j \in X_j \subset O$.
So, for any open set $O \in X$ which contains $x$ there is $i$ such that for $j \ge i$ then $x_j \in O$, which is the condition that $(x_i)$ converges to $x$.
But $V$ is open and $f(x) \in V$ and for all $i$, $f(x_i) \not \in V$, i.e. $(f(x_i))$ doesn't converge to $f(x)$.
So if $f$ is not continuous then it is not sequentially continuous and the result follows.
In 1st countable spaces sequential continuity implies continuity.
In general, net continuity implies continuity.
Here is an example which gives a very resounding no answer.
In the space $X=\beta\mathbf N\setminus \bf N$ is compact Hausdorff and all convergent sequences are eventually constant. It easily follows that every function $X\to Y$ is sequentially continuous, even though $X$ is not discrete, so there are lots of discontinuous functions from $X$.
To obtain a characterisation of continuity in terms similar to sequential continuity, you have to use nets. It is a fairly standard exercise to show that a function $f\colon X\to Y$ is continuous if and only if for each net $(x_i)_i$ in $X$ convergent to some $x$, the net $(f(x_i))_i$ is convergent to $f(x)$.
In fact, if $Y$ is $T_1$, then it is enough to assume that $f$ maps convergent nets to convergent nets (preservation of limits follows automatically), see this post.
Here's a more direct proof using the very basic definitions of first-countability and sequential continuity.
$\textbf{Definition 1 (first-countable space):}$ a topological space $(X, \mathcal{T}_X)$ is first-countable if:
$\forall x \in X$, there exists a countable sequence of (open) neighborhoods $N_1, N_2,...$ containing $x$ such that for all neighborhood $M$ of $x$ (i.e. $x \in M$), we have $N_i \subseteq M$ for some $i \in \mathbb{N}$.
The sequence $\{N_i\}_{i=1}^\infty$ is called the local basis of $x$.
Reminder 1: a trivial fact that we assume from now on is that the local basis consists of open sets.
Next we recall the definition of sequential continuity and (general) continuity:
$\textbf{Definition 2 (sequential continuity):}$ a function $f: X \to Y$ that maps between topological spaces $(X, \mathcal{T}_X)$ and $(Y, \mathcal{T}_Y)$ is sequentially continuous at $x \in X$ if $$\{x_n\}_{n = 1}^\infty \to x \quad \Longrightarrow \: \{f(x_n)\}_{n = 1}^\infty \to f(x)$$ where the LHS is a convergence in $(X, \mathcal{T}_X)$, whereas RHS in $(Y, \mathcal{T}_Y)$. This is also denoted as $\lim_{n \to \infty} f(x_n) = f(\lim_{n \to \infty}x_n) = f(x)$.
Reminder 2: the concept of convergence for general topological space (e.g. $(X, \mathcal{T}_X)$) is that for every open set $U \in \mathcal{T}_X$ that contains $x$, $\exists N > 0$ that $\forall n \geq N, x_n \in U$.
$\textbf{Definition 3 (general continuity):}$ use the same notations as above. We say that $f$ is (generally) continuous if for every open set $U \in \mathcal{T}_Y$, we have the preimage $f^{-1}(U) = \{x \in X: f(x) \in U\} \in \mathcal{T}_X$ an open set in $X$.
Now we are finally ready to ⛵ embark on the proof~
Proof sketch:
We want to take some $U \in \mathcal{T}_Y$ and show that $V := f^{-1}(U) \in \mathcal{T}_X$. To achieve this, we consider any $y \in U$ and want to find some open neighborhood $V_y$ such that $f^{-1}(y) \subseteq V_y \subseteq V$. Showing this for any $y$ then gives us an argument to show that $V$ is open.
Using the notation above, we assume $U \neq \emptyset$ and pick $y \in U$. Then take $x \in f^{-1}(y)$ -- again the case when $f^{-1}(y) = \emptyset$ is trivial.
Since $(X, \mathcal{T}_X)$ is first-countable, $x$ admits a local basis $\{N^{(x)}_n\}_{n=1}^\infty$. We now claim that there exists some $0 < m_x \in \mathbb{N}$ such that $$ x \in \bigcap_{i=1}^{m_x} N^{(x)}_i \subseteq V := f^{-1}(U) $$ i.e. a finite intersection of the local basis of $x$ lies within $V$. This claim is proved using contradiction in the next step.
By contradiction, we have that $\forall n \in \mathbb{N} > 0$, there exists some $x_n$ satisfying $$ x_n \in \bigcap_{i=1}^{n} N^{(x)}_i \quad \text{and} \quad x_n \notin V $$ Note that the first condition is well-defined since at least $x$ itself lies in any such intersections.
Also please verify yourself that such construction for $x_n$ gives us a convergent sequence $\{x_n\}_{n=1}^\infty \to x$.
Hint: it's easy to see that for any $N^{(x)}_k$ in the local basis, we have $x_n \in N^{(x)}_k$ whenever $n \geq k$. Try to generalize this to any open set $\mathcal{O} \in \mathcal{T}_X$.
Now, by the sequential continuity of $f$, we have $\{f(x_n)\}_{n=1}^\infty \to f(x) = y$.
⚠️Contradiction occurs since by construction, $\forall n$, we have $x_n \notin f^{-1}(U)$, so $f(x_n) \notin U \ni y$.
We thus prove the claim in Step 1.
Let's reflect on what we have done so far: for any $y \in U$ and $x \in f^{-1}(y)$, we show that $$ x \in M^{(x)} := \bigcap_{i=1}^{m_x} N^{(x)}_i \subseteq V := f^{-1}(U) $$ where each $M^{(x)}$ is a finite intersection of open sets, thus open. Taking the union for all $x \in V$: $$ V \subseteq \bigcup_{x \in V}M^{(x)} \subseteq V $$ which essentially tell us that $V = \bigcup_{x \in V}M^{(x)}$. Since any union of open sets is open, $V$ is open too.
This essentially wraps up the proof since $U$ is any open set in $\mathcal{T}_Y$ and its preimage is open with $f$.
Continuity implies sequential continuity in all cases. Sequential continuity implies continuity in second-countable spaces.
If $f: X \rightarrow Y$ is continuous, then if $x_n \rightarrow x$, given a neighborhood $V$ of $f(x)$, by continuity, $f^{-1}(V)$ is a neighborhood of $x$, and $x_n \in f^{-1}(V)$ for $n \geq n_0$. But this implies $f(x_n) \in V$ for $n \geq n_0$, and therefore $f(x_n) \rightarrow f(x)$.
On the other hand, if we are in a space with a countable base, and $f$ is sequentially continuous but not continuous, then there exists some open set $V$ such that $f^{-1}(V)$ is not open. For each $x \in f^{-1}(V)$, if $x$ is in some open set $U_x$ with $f(U_x) \subseteq V$, then $f^{-1}(V)$ would be open as a union of these open sets $U_x$ with varying $x$. Therefore, there must exist some $x \in f^{-1}(V)$ for which no open neighborhood $U_x$ has $f(U_x)$ entirely contained in $V$ (Lemma 1). Let $\{B_j(x)\}_{j=1}^\infty$ be basis open sets of $X$ containing $x$. Define $B_J(x) := \cap_{j=1}^J B_j(x)$, which are also open sets. By Lemma 1, for each $B_J$, there exists $b_J \in B_J(x)$ such that $f(b_J) \in V^C$. Clearly, $b_J \rightarrow x$ by construction, but $f(b_J) \not\rightarrow f(x) \in V$. This contradicts sequential continuity and completes the proof.
