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 $f:X\to X$?

  • $\begingroup$ 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 $X$. $\endgroup$ – Damien Aug 5 '11 at 19:36
  • 1
    $\begingroup$ How about the line with two origins? $\endgroup$ – Mark Aug 5 '11 at 19:59
  • $\begingroup$ @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 $X$. $\endgroup$ – JSchlather Aug 5 '11 at 20:17
  • $\begingroup$ @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. $\endgroup$ – Jason DeVito Aug 5 '11 at 20:17
  • 2
    $\begingroup$ 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. $\endgroup$ – 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.

  • 3
    $\begingroup$ I don't understand the last step. What about a sequence of rational approximations to $\sqrt{2}$? $\endgroup$ – Qiaochu Yuan Aug 5 '11 at 22:01
  • 4
    $\begingroup$ @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) $\endgroup$ – Sam Aug 5 '11 at 22:16
  • 3
    $\begingroup$ In Case I, why must $f|_\mathbb{Q}$ take values only in $\mathbb{Q}$? $\endgroup$ – user83827 Aug 5 '11 at 22:53
  • 4
    $\begingroup$ @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. $\endgroup$ – user83827 Aug 6 '11 at 0:40
  • 2
    $\begingroup$ @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. $\endgroup$ – 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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.