# If every continuous $f:X\to X$ has $\text{Fix}(f)\subseteq X$ closed, must $X$ be Hausdorff?

Given a function $f:X\to X$, let $\text{Fix}(f)=\{x\in X\mid x=f(x)\}$.

In a recent comment, I wondered whether $X$ is Hausdorff $\iff$ $\text{Fix}(f)\subseteq X$ is closed for every continuous $f:X\to X$ (the forwards implication is a simple, well-known result). At first glance it seemed plausible to me, but I don't have any particular reason for thinking so. I'll also repost Qiaochu's comment to me below for reference:

I would be very surprised if this were true, but it doesn't seem easy to construct a counterexample. Any counterexample needs to be infinite and $T_1$, but not Hausdorff, and I don't have good constructions for such spaces which don't result in a huge collection of endomorphisms...

Is there a non-Hausdorff space $X$ for which $\text{Fix}(f)\subseteq X$ is closed for every continuous $f:X\to X$?

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If $f,g: X \to Y$ where $Y$ is $T_2$, and $f,g$ are continuous then $\{x: f(x) = g(x) \}$ is closed in $X$. –  Damien Aug 5 '11 at 19:36
How about the line with two origins? –  Mark Aug 5 '11 at 19:59
@Mark, I don't think so. I'm fairly sure that you can continuously map the line with two origins to itself, but map both of the origins $0_a,0_b$ to $0_a$. To see this consider the line with two origins $X$ as a quotient of $Y:=\mathbb R \times \{a\} \sqcup \mathbb R \times \{b\}$, $p:X \rightarrow Y$. Then the map projecting $Y$ onto $\mathbb R \times \{a\}$ respects the identifications of $p$ so it induces a continuous map $f: Y \rightarrow X$ and then $p \circ f$ is the desired map. Note then that $fix(p\circ f)$ is actually an open subset of $X$. –  JSchlather Aug 5 '11 at 20:17
@Mark: If I'm not mistaken, for the function $f$ which is the identity except for swapping the two origins, $Fix(f)$ is not closed. –  Jason DeVito Aug 5 '11 at 20:17
I suspect that if one took a Cook continuum (a continuum with only trivial self-maps), fixed a point, and weakened the topology at that point to the cofinite topology, one would have an example, but I’d need to see the construction of a Cook continuum to be sure. One could probably start with just about any strongly rigid space, in fact. Unfortunately, I don’t have any of the constructions handy to check. –  Brian M. Scott Aug 5 '11 at 21:37

Let me propose the following counterexample:

Take $X = \overline{\mathbb Q}$ the one-point compactification of $\mathbb Q$. This space is not Hausdorff, since $\mathbb Q$ is not locally compact (the problem is $\infty$).

Now let $f: \overline{\mathbb Q} \to \overline{\mathbb Q}$ be a continuous function, and let $x\in \overline{\mathrm{Fix}(f)}$ be an arbitrary point in the closure of $\mathrm{Fix}(f)$.

Case I: Suppose $\infty \in \mathrm{Fix}(f)$. Then either $x = \infty$ or we must have that the restriction $f|_{\mathbb Q}: \mathbb Q \to \overline{\mathbb Q}$ is continuous. But then also $x \in \overline{\mathrm{Fix}(f|_\mathbb{Q})} \subset \mathrm{Fix}(f|_\mathbb{Q}) \cup \{\infty\}= \mathrm{Fix}(f)$.

Case II: Now suppose $\infty \notin \mathrm{Fix}(f)$ and $x\ne \infty$. Then there is a convergent sequence $x_n \to x$ with $x_n\in \mathrm{Fix}(f)$. But then by continuity: $f(x) = \lim_{n\to \infty} f(x_n) = \lim_{n\to \infty} x_n = x$.

So there is only one case left:

Can we have $x=\infty \in \overline{\mathrm{Fix}(f)}$ but at the same time $\infty \notin \mathrm{Fix}(f)$?

If this were the case, then $\mathrm{Fix}(f)$ would definitely not be compact. But this implies that there must be a sequence in $x_n \in \mathrm{Fix}(f) \subset \mathbb Q$ without a convergent subsequence - that is: no convergent subsequence in $\mathbb Q$!

But then such a sequence has a subsequence converging to $\infty$, which implies that $\infty \in \mathrm{Fix}(f)$.

Hoping I haven't made some silly mistake, this concludes the argument that this $X$ is indeed a counterexample.

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I don't understand the last step. What about a sequence of rational approximations to $\sqrt{2}$? –  Qiaochu Yuan Aug 5 '11 at 22:01
@Qiaochu: Let $U$ be any open neighborhood of $\infty$: By definition $C = \mathbb Q \setminus U$ is compact. Now $C$ has positive distance to $\sqrt 2$, from which it follows that $x_n \notin C$ (or equivalently $x_n \in U$) for almost all $n$. Therefore $x_n \to \infty$. (In fact, $\overline{\mathbb Q}$ is sequentially compact, as can be shown using the above) –  Sam Aug 5 '11 at 22:16
In Case I, why must $f|_\mathbb{Q}$ take values only in $\mathbb{Q}$? –  user83827 Aug 5 '11 at 22:53
@Sam: You're welcome. I think I understand what's going on now. All you need is any sequential, non-Hausdorff space with unique limits. Then the (unique) limit of a sequence of fixed points is again fixed! Good idea. –  user83827 Aug 6 '11 at 0:40
@Harry: Any "traditionally" convergent sequence in $\mathbb{Q}$ is contained in a compact subset of $\mathbb{Q}$, and thus can be separated from the point at infinity. –  user83827 Aug 6 '11 at 3:52

While I think that Sam's counterexample is fine, I would like to elaborate on ccc's comment to Sam's answer.

Recall that a topological space $X$ is called Fréchet if for every $A\subseteq X$ and every point $x\in \bar{A}$ there exists a sequence of points of $A$ converging to $x$.

Observe that if $X$ is Fréchet with unique sequential limits, then $\text{Fix}(f)$ is closed for every continuous $f:X\to X$. Indeed, if $X$ is such a space, $f:X\to X$ is continuous and $x\in\overline{\text{Fix}(f)}$, then by Fréchetness of $X$ we find a sequence $(x_n)_{n=1}^\infty$ of points of $\text{Fix}(f)$ converging to $x$. Then $f(x)=f(\lim_{n\to\infty} x_n)=\lim_{n\to\infty}f(x_n)=\lim_{n\to\infty}x_n=x$ by continuity of $f$ and uniqueness of sequential limits in $X$. This means that $x\in\text{Fix}(f)$ and hence $\text{Fix}(f)$ is closed.

To give a positive answer to the question we want to find a non-Hausdorff Fréchet space with unique sequential limits. The one-point compactification of the rationals suggested by Sam is such a space. Another example is Example 6.2 in S.P. Franklin, Spaces in which sequences suffice II, Fund. Math. 61 (1967), which is available here.

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This is a remark related to the question.

Theorem : let $X$ be a topological space. Suppose that for every space $Y$ and pair of maps $f,g : Y \rightarrow X$, the subset $\{ y \in Y : f(y)=g(y) \}$ is closed. Then $X$ is Hausdorf.

It is easy to prove by considering the $2$ projections $X \times X \rightarrow X$. In algebraic geometry, a (pre)variety which satisfies this axiom is called separated (see for example Milne's note on page 61 http://www.jmilne.org/math/CourseNotes/ag.html). This notion was introduced because algebraic varieties are not Hausdorff, and if we work with separated varieties, then maps are determined on dense subset.

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