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 $X$.


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.

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.

  • $\begingroup$ I removed my comment it was wrong! Sorry! $\endgroup$ Nov 11, 2010 at 10:34
  • $\begingroup$ Minor point: In your $T_1$ case, your set $E=\{a\}$ is not dense (in fact, it's closed). You should change $E$ to $\{b\}$. This also means you have to modify $f$ and $g$. The function $f$ should satisfy $f(a) =u$ and $f(b) = v$, while $g$ should satisfy $g(a)=g(b) = v$. $\endgroup$ Nov 11, 2010 at 12:36
  • $\begingroup$ @Jason De Vito: Thanks; I had it right originally, then screwed up the write up. $\endgroup$ Nov 11, 2010 at 15:49
  • $\begingroup$ @Arturo: "As a result of this question, I've been thinking about the following ...". Please explain the relation between the question given in the hyperlink "this question" and the current question. Its baffling to me ! $\endgroup$
    – Rajesh D
    Nov 11, 2010 at 16:07
  • 1
    $\begingroup$ @Rajesh D: Technically, you are asking me to violate copyright and the Terms of Use of JSTOR. I'll e-mail you a summary of the problem/solution instead. $\endgroup$ Nov 15, 2010 at 5:09

3 Answers 3


[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 $(Y, T)$.
  • 2
    $\begingroup$ Thank you for moving it here. So, it looks to me like this is the "fix" to my argument; I just needed to take a slightly finer topology on $X$ than I was taking (though your argument I think is better because it looks like it does not require the Axiom of Choice). I'll let it stand for a bit and think through your equivalent statements before accepting, but it looks good! Thank you. $\endgroup$ Nov 13, 2010 at 4:53

This problem has a very straightforward solution when you conceptualize convergence of nets in terms of continuity of maps. Given a directed set $I$, the statement that a net $(y_i)_{i\in I}$ in a space $Y$ converges to a point $y$ can be expressed in terms of continuity of a map. Namely, let $X=I\cup\{\infty\}$, topologized such that a set is open iff either it does not contain $\infty$ or it contains the set $\{j\in I:j\geq i\}$ for some $i\in I$. Then $(y_i)$ converges to $y$ iff the map $i\mapsto y_i$, $\infty\mapsto y$ is a continuous map from $X$ to $Y$.

Now we use the characterization of Hausdorff spaces in terms of nets: a space is Hausdorff iff every net has at most one limit. So if $Y$ is not Hausdorff, there is some net $(y_i)_{i\in I}$ in $Y$ which converges to two distinct points $y$ and $y'$, which means that the map $I\to Y$ sending $i$ to $y_i$ can be extended continuously to $X$ by sending $\infty$ to either $y$ or $y'$. Since $I$ is dense in $X$, this is exactly what we're looking for.


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.

  • 1
    $\begingroup$ Thanks for this! I can't spare much time right now, and it's been over 15 years since the last time I did any serious thinking about nets, so I need to come back and think about the idea of using nets to show $g$ is continuous. But in some sense it looks promising, since the entire definition of $X$ has to do with nets; I was having trouble relating back to the construction of $X$ when trying to show $g$ is continuous. For the cofinite topology I can prove the process "works", and also for an interval with a doubled point. But those may not be "generic enough" the way Sierpinski is. $\endgroup$ Nov 11, 2010 at 20:38
  • $\begingroup$ It occurs to me that "any nonconstant net that converges to s or t converges to both s and t" might be too strong. But you don't really need that. You just need one net in E converging to both s and t. $\endgroup$
    – kahen
    Nov 12, 2010 at 0:55

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