If $f$ is a contraction map and $X$ is compact, show that there exists a fixed point [duplicate]

Question

Let $(X, d)$ be a compact metric space, and $f: X \to X $ is a map such that $d (f(x), f(y)) < d(x, y)$ where $x \neq y$. Show that there exists a point such that $f(x) = x$.

My idea is to make compact sets where any finite intersection is non-empty and then use the fact that an infinite intersection will also not be empty to show existence. I don't want to use continuity as that has not been covered in my notes yet.

Thus take $A_n = \{x \in X:d(f^{n-1}(x), f^n(x)) < a \}$, where $f^n$ is $f$ iterated $n$ times. Since $A_n \subset X$, they are compact and any finite intersection is non-empty. Thus an infinite intersection is also non-empty.

Since $a$ was arbitrary and distances are always positive, I can create a series of non-empty $A_n$'s between $0$ and $a$ no matter how small $a$ is. Thus there exists a point where $f(x) = x$.

How do I know that you can always find points with distance less than any $a$? Because $X$ is compact, it is bounded and so there is a maximum distance between any 2 points. So start there, iterate $f$ once and now you have points with distance less than $a$. Repeat as needed.

I feel like it's missing some rigor though, or at least could use some clean-up or an extra piece that I've overlooked.

Answer 1

There's a number of issues here. The first one is that you claim that your sets $A_n$ are compact, but you provide no proof. There are two issues with this: firstly, you used a strict inequality $$d(f^{n-1}(x),f^n(x))<a$$ and the set of $x$ that satisfy this is not necessarily closed - for instance, if $f(x)=\frac{x}2$ and $n=1$ and $a=1$, then this set is $(-4,4)$, which isn't compact. In metric spaces, a subset of a compact space is compact if and only if it is closed, so this is a problem. Moreover, if your idea is to avoid using that $f$ is continuous, you have a problem here - we need to know that the map $x\mapsto d(f^{n-1}(x),f^n(x))$ is continuous before we can do much reasoning about the topological properties of sets like $A_n$, which may be expressed as preimages thereof.

Another issue is that you actually have an increasing sequence of $A_n$'s, so in particular, every $A_n$ contains $A_1$, and the intersection of all of them is $A_1$. This is probably not what you wanted. It is also not clear to me how you are reasoning that any finite intersection is non-empty (or even how you are reasoning that any $A_i$ alone is non-empty). The limiting argument that you try to use afterwards to find such an $x$ is very unclear, since it seems like you want to construct these sequences for varying $A_n$ - but you never specify what set or intersection $x$ is coming out of.

I think the heart of your argument is the last paragraph - one can certainly prove this by first finding two a point which is separated from its image by some finite distance, then "iterating" to find points that are closer to their image than the last pair. However, "repeat as necessary" isn't really enough, since if we get a sequence of distances like $2,\,1+\frac{1}2,\,1+\frac{1}3,\,1+\frac{1}4,\ldots$, we repeat infinitely often and never actually get to a point and its image being a distance of $0$ apart.

I really doubt that this can be proven without using as a lemma that $f$ is continuous, which the given $d(f(x),f(y))<d(x,y)$ certainly implies. There are a number of ways to repair the proof that you have, but it needs a lot of work to be made valid. One direction that I think this is trying to go is to consider the sequence of compact sets $X,\,f[X],\,f^2[X],\,f^3[X],\ldots$. It can be shown that any point in the intersection of all of those sets is a fixed point for an $f$ that has the given property, though to show that these are compact sets and to finish the argument that points in their intersection have the desired property, its continuity is crucial.