Characterization of Fréchet-Urysohn spaces using sequential continuity at a point

Question

A map $f \colon X \to Y$ is called sequentially continuous at the point $a$ if for every sequence $(x_n)$ such that $x_n\to a$, we also have $f(x_n)\to f(a)$. $$x_n\to a \qquad \Rightarrow \qquad f(x_n) \to f(a).$$ A map $f \colon X \to Y$ is called sequentially continuous, if it is sequentially continuous at each point of $X$.

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In sequential spaces sequential continuity and continuity are equivalent. In fact, this property characterizes sequential spaces in the sense that if this equivalence holds for every topological space $Y$ and for every map $X\to Y$, then $X$ is sequential. This characterization is mentioned also in Wikipedia article. Although I am not able to find a reference for this fact in a textbook or an article at this moment. (I'd say this is well-known and it can be considered folklore.)

One possible way to see this: If $X$ is not sequential and $sX$ is the sequential coreflection of $X$, then $id_X \colon X \to sX$ is sequentially continuous but not continuous. By sequential coreflection $sX$ of $X$ I mean the topological space which has the same underlying set as $X$ but closed sets are only those sets, which are sequentially closed in the original topology.

A different argument: This can also be seen from the fact that a sequential space $X$ can be obtained as a quotient space of a topological sum, where summands correspond to convergent sequences in $X$. (See, for example, proof of Theorem 3.10 in [G].) A map from a quotient space is continuous if and only if the composition with the quotient map is continuous. Continuity of the composition map is precisely convergence of the image of corresponding sequence.

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What I am interested in is the following characterization of Fréchet-Urysohn spaces (a.k.a. Fréchet spaces): A space $X$ is Fréchet-Urysohn if and only if continuity at a point is equivalent to sequential continuity at a point. (Again, in the sense that this equivalence holds for each $x\in X$, any topological space $Y$ and for each map $f\colon X\to Y$.)

This characterization was not known to me until it came up in a discussion with my colleague recently. My main question is:

• Do you happen to know about any reference for this characterization of Fréchet-Urysohn spaces?

But I'd be grateful for providing alternative proofs to the one I suggested below, too. (Although that was not the main reason for posting this question.)

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I'll add sketch of the proof. I do not claim that this it the most direct proof, but for some reasons the remaining equivalent conditions were interesting for me, too.

The following conditions are equivalent for a topological space $X$:
(a) $X$ is Fréchet-Urysohn.
(b) $X$ is hereditary sequential (i.e., every subspace of $X$ is sequential).
(c) For each $a\in X$ the prime factor $X_a$ is Fréchet.
(d) For each $a\in X$ the prime factor $X_a$ is sequential.

If $a\in X$, then the prime factor $X_a$ of $X$ at $a$ is the space which has the same neighborhoods of $a$ as $X$ and all points other than $a$ are isolated. I have seen this terminology in several papers, see references in my answer here. Thomas Andrews called it local topology in one of his recent answers.

The important fact is that $f \colon X_a \to Y$ is continuous if and only $f \colon X\to Y$ is continuous at $a$.

Sketch of a proof. (a) $\Leftrightarrow$ (b): This is well-known, see [F, Proposition 7.2], [E, Exercise 2.4.6].

(a) $\Rightarrow$ (c): We want to show that if $a\in\overline V$ holds in $X_a$, then there is a sequence of points of $V$ which converges to $a$. But, since the neighborhoods of $a$ in $X_a$ and in $X$ are the same, $a\in\overline V$ holds in the space $X$, too. Since this space is Fréchet, there is a sequence of points of $V$ converging to $a$ in $X$. This sequence converges to $a$ in $X_a$, too.

(c) $\Rightarrow$ (d): Obvious.

(d) $\Rightarrow$ (b): We can use the observation that every space can be obtained as a quotient of the topological sum of all its prime factors. This shows that $X$ is sequential. If $Y$ is a subspace of $X$ and $a\in Y$ then we have a quotient map $X_a\to Y_a$ (simply by mapping points from $Y\setminus X$ to the point $a$). Using the same argument again, we get that $Y$ is sequential.

[E] R. Engelking, General Topology. Sigma Series in Pure Mathematics 6. Berlin: Heldermann, revised ed., 1989.
[F] S.P. Franklin, Spaces in which sequences suffice II, Fund. Math. 61 (1967); eudml.
[G] A. Goreham: Sequential convergence in topological spaces, http://arxiv.org/abs/math/0412558

Answer 1

I'd like to add a more direct proof:

1. • Let X be a Fréchet-Urysohn space and $f : X \to Y$ be continuous at $x \in X$. Then for each neighborhood $U$ of $f(x)$, there is a neighborhood $V$ of $x$, such that $f(V) \subset U$. Given a sequence converging to $x$, all elements of it, beginning from some $x_n$, lie in $V$, so respective $f(x_n)$ lie in $U$. Since this holds for any neighborhood $U$, $f(x_n)$ converges to $f(x)$ (we don't really need the Fréchet-Urysohn condition for that case).
   • Let X be a Fréchet-Urysohn space, we will prove the result by contraposition: if $f:X \to Y$ is not continuous at $x$, then it must be not sequentially continuous at $x$ either. Since $f$ is not continuous at $x$, there is an open neighborhood $U$ of $f(x)$, such that every neighborhood $V$ of $x$ does not lie completely in $f^{-1}(U)$. Consider $\overline{X\setminus f^{-1}(U)}$. Every neighborhood of $x$ has a point not lying in $f^{-1}(U)$, therfore $x\in\overline{X\setminus f^{-1}(U)}$. Since $X$ is a Fréchet-Urysohn space, sequential closure $scl (X\setminus f^{-1}(U)) = \overline{X\setminus f^{-1}(U)}$, so, there is a sequence $x_n$, converging to $x$ and lying completely in $X\setminus f^{-1}(U)$, thus $\forall n\in\mathbb{N}\ f(x_n)\notin U$; By definition of limit $f(x_n)\nrightarrow f(x)$, so $f$ is not sequentially continuous at $x$ either.
2. • We want to prove, that if for any topological space $Y$ continuity of $f: X\to Y$ is equivalent to sequential continuity at that point, then $X$ is a Fréchet-Urysohn space. I am going to write down, how I came up with the proof. You can scroll to the next paragraph, if you only want the proof. I tried to understand, what is exactly the reason we are using the Fréchet-Urysohn condition for. We prove, that if $f$ is not continuous at $x$, then it is not sequentially continuous at $x$. So, we must consider some function that is not continuous at $x$ and use, that there is a sequence $x_n\rightarrow x$ with some properties. Since we need to prove, that any point of $\overline{A}\setminus A$ belongs to $scl A$, we need $x_n$ to lie in A and x to lie in $\overline{A}\setminus A$. We need to find some closed set with $x$ in it, that is related to discontinuity of $f$ at $x$. We have already seen one such set: $\overline{X\setminus f^{-1}(U)}$. So, we try to choose $f$, so that for given $A$, $A=X\setminus f^{-1}(U)$. If $x\in\overline{X\setminus f^{-1}(U)}$ for $U$, an open neighborhood of f(x) in $Y$, then any neighborhood of $x$ is not lying completely in $f^{-1}(U)$; so $f$ is discontinuous at $x$, and there is a sequence $x_n\rightarrow x$, such that $f(x_n)\nrightarrow f(x)$, or equivalently, there is a neighborhood $W$ of $f(x)$, such that there are infinitely many elements of $x_n$ with $f(x_n)\notin W$. Therefore, there are infinitely many elements of $x_n$ with $f(x_n)\notin W\cap U$. If we relabel $x_n$ as the sequence of such elements, then we get: $x\in scl(X\setminus f^{-1}(U\cap W))$. If $U\cap W$ would be equal $U$, then that would finish the proof. Thus, we would benefit, if we had a topology on $Y$ with $U$ being the smallest neighborhood of $f(x)$. Let's try to find such $f$ and topological space $Y$, that $A=X\setminus f^{-1}(U)$, $U$ is open and $f(x)\in U$(which is satisfied automatically) and $U$ is the smallest neighborhood of $f(x)$. We can consider the situation, where $U$ and $Y\setminus U$ are open one-point sets, then all the conditions are satisfied.
   • Given a set $A\subset X$, we want to prove, that $scl A = \overline{A}$. Consider indicator function $1_{A}: x\mapsto \begin{cases} 1,& \text{if } x\in A\\ 0, & \text{otherwise} \end{cases}$. $Y = \{0, 1\}$ with discrete topology. Given a point $x\in\overline{A}\setminus A$, $x$ is a boundary point of $A$, so every neighborhood of $x$ has not empty intersection with both $A$ and $X\setminus A$, therefore $f$ is not continuous at $x$: $f(x)=0$, $\{0\}$ is an open neighborhood of $0$, but every open neighborhood of $x$ has points $y$, such that $1_A(y)=1$. Therefore, $f$ is not sequentially continuous at $x$, so there is a sequence $x_n \rightarrow x$, such that $f(x_n)\nrightarrow f(x)$, thus there are infinitely many elements of the sequence lying in $A$ (otherwise, there are only finitely many elements of $f(x_n)$ not equal to $f(x)=0$), so we can form a subsequence, converging to $x$, as a subsequence of a sequence converging to $x$ and lying completely in $A$. Therefore $x\in sclA$. This finishes the proof, since $A\subset sclA\subset\overline{A}$ and we have proven that $\overline{A}\setminus A\subset sclA$