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 $\mathbb{R}.$


5 Answers 5


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.

  • $\begingroup$ sequentially closed sets are closed, is the definition. The continuity part then follows. See my answer. $\endgroup$ Jul 2, 2017 at 6:14
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    $\begingroup$ In the proof that seq. continuity over a 1-countable space implies continuity it seems to me that you cannot use 1-countability directly to conclude that $f^{-1}(U)$ contains an open set. 1-countability only tells you that you have a countable neighborhood basis for $x$, so you first need to prove that $f^{-1}(U)$ is a neighborhood of $x$...which, by def., means that it contains an open set! I'd suggest to go by contradiction: assume it is not and use 1-countability to define a sequence converging to $x$ but not contained in $f^{-1}(U)$, which contradicts the seq. continuity of $f$. $\endgroup$
    – Manlio
    May 13, 2018 at 10:39
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    $\begingroup$ I am having a few doubts regarding $\mathbb{R}$ over $\mathbb{Z}$ $\endgroup$ Sep 4, 2020 at 15:01
  • $\begingroup$ It is not true in general that $f(f^{-1}(U)) = U$. $\endgroup$ Aug 11, 2022 at 14:03

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
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.


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