Existence of fixed point for a continuous function on an infinite closed set

Question

Why is it that a continuous surjection from $X\to X$ has a fixed point when $X=[1,2]\cup[3,\infty)$ and $X=[3,\infty)$ but not when $X=[1,2]\cup[3,7]$?

When $X=[3,\infty)$ , since the set is closed and connected, therefore the image should also be connected, whence it seems intuitive to expect a fixed point. But, how does $[1,2]\cup[3,\infty)$ have a fixed point but $[1,2]\cup[3,7]$ does not? Any rigorous reasoning? Thanks beforehand.

Answer 1

Closed and connected just isn't enough on its own to produce a fixed point. The space $X = (-\infty,+\infty)$ is closed and connected, but the continuous surjection $f(x)=x+1$ has no fixed point.

Instead, the topological space $X = [1,2] \cup [3,\infty)$ has some explicit features which force it to have the fixed point property:

1. $X$ has two components;
2. One component $[1,2]$ is compact and has the fixed point property; the other component $[3,\infty)$ is noncompact.

From this you can deduce the following for any continuous self-surjection $f : X \to X$:

• The map $f$ induces a surjection of the components of $X$ (otherwise $X$ has a component disjoint from the image of $f$, contradicting surjectivity).
• $f$ induces a bijection of the components of $X$ (because a surjection from a finite set to itself is a bijection).
• $f$ takes each component surjectively to the image component (otherwise that image component is not contained in the image of $f$).
• $f$ must take $[1,2]$ to itself (because the continuous image of a compact set is compact so $f$ cannot take $[1,2]$ surjectively to $[3,\infty)$).
• $f$ has a fixed point on $[1,2]$ (because $[1,2]$ has the fixed point property).

Try tracing through this argument for $X = [1,2] \cup [3,7]$ to see how it breaks down.

The case of $X=[3,\infty)$ is a different story altogether. It does indeed have the fixed point property, but by a different argument.

Regarding another example $X = [1,2] \cup [3,7)$: that $X$ is homeomorphic to $[1,2] \cup [3,\infty)$, and therefore that $X$ does have the fixed point property (or, if one doesn't like the homeomorphism argument, you can just trace through the exact same argument that I gave for $[1,2] \cup [3,\infty)$).

If there's a moral to this story, it's that the fixed point property can be true for many different reasons.

Answer 2

A priori, we cannot ever expect a fixpoint for all continuous $X\twoheadrightarrow X$ unless $X$ has suitable special properties.

The problem we encounter with $X=[1,2]\cup [3,\infty)$ is that it is not compact; specifically, it is the direct sum (i.e., disjoint union) of a compact and a connected non-compact space. The connected image of a compact space is always compact, hence cannot cover the non-compact component; therefore it needs the non-compact component to cover itself (at least near $\infty$), which means (by connectedness) that the non-compact component cannot contribute at all to the compact component, so that the compact needs to cover itself. Then we obtain a fixpoint (in fact, two) from the fact that $[3,\infty)\twoheadrightarrow [3,\infty)$ and $[1,2]\twoheadrightarrow[1,2]$ both must have fixed points.

With $X=[1,2]\cup[3,7]$, we have the obvious counterexamples that swap the two (topologically indistinguishable) components.

Note that for example $$[3,\infty)\cup [1,2]\cup[-1,0]\cup [-3,-2]$$ also gives us a fixed point because again we need $[3,\infty)\twoheadrightarrow [3,\infty)$ (though the other components may be permuted). But with infinitely many intervals $$[3,\infty)\cup [1,2]\cup\bigcup_{n\in\Bbb N_0}[-2n-1,-2n] ,$$ we are not guaranteed a fixpoint. Then again, $$[3,\infty)\cup\bigcup_{n\in\Bbb N_0}\{-n\}$$ looks "similar", and does guarantee a fixpoint