Does this property characterize a space as Hausdorff?

Question

As a result of this question, I've been thinking about the following condition on a topological space $Y$:

For every topological space $X$, $E\subseteq X$, and continuous maps $f,g\colon X\to Y$, if $E$ is dense in $X$, and $f$ and $g$ agree on $E$ (that is, $f(e)=g(e)$ for all $e\in E$), then $f=g$.

If $Y$ is Hausdorff, then $Y$ satisfies this condition. The question is whether the converse holds: if $Y$ satisfies the above condition, will it necessarily be Hausdorff?

If $Y$ is not at least $T_1$, then $Y$ does not have the property: if $u,v\in Y$ are such that $u\neq v$ and every open neighborhood of $u$ contains $v$, then let $X$ be the Sierpinski space, $X=\{a,b\}$, $a\neq b$, with topology $\tau=\{\emptyset,\{b\},X\}$, $E=\{b\}$, let $f,g\colon X\to Y$ be given by $f(a)=f(b)=v$, and $g(a)=u$, $g(b)=v$. Then both $f$ and $g$ are continuous, agree on the dense subset $E$, but are distinct.

My attempt at a proof of the converse assumes the Axiom of Choice and proceeded as follows: assume $Y$ is $T_1$ but not $T_2$; let $u$ and $v$ be witnesses to the fact that $Y$ is not $T_2$, let $\mathcal{U}\_s$ and $\mathcal{V}\_t$ be the collection of all open nbds of $s$ that do not contain $t$, and all open nbds of $t$ that do not contain $s$, respectively. Construct a net with index set $\mathcal{U}\_s\times\mathcal{V}\_t$ (ordered by $(U,V)\leq (U',V')$ if and only if $U'\subseteq U$ and $V'\subseteq V$) by letting $y_{(U,V)}$ be a point in $U\cap V$ (this is where AC comes in). Let $E=\{y_{(U,V)}\mid (U,V)\in\mathcal{U}\_s\times\mathcal{V}\_t\}$, and let $X=E\cup\{s\}$. Give $X$ the induced topology; let $f\colon X\to Y$ be the inclusion map, and let $g\colon X\to Y$ be the map that maps $E$ to itself identically, but maps $s$ to $t$.

The only problem is I cannot quite prove that $g$ is continuous; the difficulty arises if I take an open set $\mathcal{O}\in \mathcal{V}_t$; the inverse image under $g$ is equal to $((\mathcal{O}\cap X)-\{t\})\cup\{s\}$, and I have not been able to show that this is open in $X$.

So:

Does the condition above characterize Hausdorff spaces?

If not, I would appreciate a counterexample. If it does characterize Hausdorff, then ideally I would like a way to finish off my proof, but if the proof is unsalvageable (or nobody else can figure out how to finish it off either) then any proof will do.

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Added: A little digging turned up this question raised in the Problem Section of the American Mathematical Monthly back in 1964 by Alan Weinstein. The solution by Sim Lasher gives a one paragraph proof that does not require one to consider $T_1$ and non-$T_1$ spaces separately.

Answer 1

[This answer moved from this question on Arturo's advice]

I think the following goes a long way towards proving a converse:

Let $(Y, T)$ be any T1 topological space with at least two points and let $a$ and $b$ be distinct points in $Y$.

Let $X = Y\setminus\{b\}$. Let $f: X \to Y$ be the inclusion of $X$ in $Y$. Let $g: X \to Y$ agree with $f$ on $X\setminus\{a\}$ and $g(a) = b$.

Finally, define the topology on $X$ to be the coarsest topology that makes both $f$ and $g$ continuous.

With these assumptions it turns out that $X\setminus\{a\}$ is dense in $X$ if and only if $a$ and $b$ do not have disjoint neighbourhoods in $Y$.

To see that this is true, let us construct a base of the topology on $X$. To make $f$ continuous we only need to take the subspace topology. Since X is open in Y this is $S_1 = \{ G \in T \mid b \notin G \}$. To also make $g$ continuous we need to add the open neighbourhoods of $b$, with $b$ replaced by $a$. This gives $S_2 = \{ (H\setminus\{b\} \cup \{a\} \mid H \in T, b \in H \}$. Now $S_1 \cup S_2$ is a subbase of the topology on $X$. Since $S_1$ and $S_2$ are already closed under finite intersection, and each covers $X$, we can say that $B = \{ G \cap H \mid G \in S_1, H \in S_2 \}$ is a base of the topology.

Then (remembering that finite sets are closed in Y) we find that the following are all equivalent:

• $X\setminus\{a\}$ is not dense in $X$
• $\{a\}$ is open in $X$
• $\{a\} \in B$
• there are $G \in S_1, H \in S_2$ such that $G \cap H = \{a\}$
• there are $G \in S_1, H \in S_2$ such that $a \in G$ and $G \cap (H\setminus\{a\}\cup\{b\}) = \oslash$
• $a$ and $b$ have disjoint neighbourhoods in $(Y, T)$.

Answer 2

Here's a sketch of my idea for proving that $g$ is continuous:

1. Show that $s$ and $t$ are witnesses that $Y$ is not $T_2$ iff any eventually nonconstant net converging to $s$ or $t$ converges to both $s$ and $t$.
2. Observe that $g$ is continuous on $E$ (why?).
3. Show that $g$ is continuous at $s$ by using the convergent net definition as follows: Let $(x_\alpha)$ be a net that converges to $s$. We split into cases:
   1. $(x_\alpha)$ is eventually constant. Here we need to use that $Y$ is $T_1$. Because of this, $x_\alpha$ is eventually equal to $s$, so $g(x_\alpha)$ is eventually equal to $t$ and hence $g(x_\alpha) \to t = g(s)$.
   2. $(x_\alpha)$ is not eventually constant and converges to $s$. By the first subresult, we have $g(x_\alpha) \to t = g(s)$.

(NB! My topology class used Munkres' "Topology" which outside of a few exercises never mentions nets. So it might be wrong and full of holes to fill out)

I spent a lot of time trying to figure out some cleverer $X$ than the one you came up with, but it was really hard. It seems obvious though that it has to be constructed from $Y$ somehow. Some of my ideas involved starting with $2^Y$, i.e. the set of indicator functions on $Y$, giving it the product topology and identifying with $\mathcal{P}(Y)$ and from there you can pass to a topology on $E = \lbrace U \cap V \mid U \in \mathcal{U}_s, V \in \mathcal{V}_t \rbrace$, but that effectively boils down to what you did. It doesn't even avoid choice as you still have to use that to define $f$.

Another idea I tried was to consider only the case where $Y$ is infinite and then constructing something using the cofinite topology, but that too didn't pan out.