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Let $(X,d)$ be a compact metric space. Let $f:X\to X$ be such that $d(f(x),f(y))<d(x,y)$ for all $x,y\in X$ but $x\neq y$. Can $f$ be surjective?

I (believe) I've shown $f$ is always continuous and always has a fixed point. But, I cannot seem to pin down whether it can be onto. I don't know if there may be some exotic compact space that would allow so, but my intuition says the answer is 'no.'

Continuous: Simply choose $\delta=\epsilon$.

Fixed point: Proof by contradiction using $x\mapsto d(x,f(x))$.

Surjective: (?)

PrincessEev
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user12386
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2 Answers2

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Given a metric space $(X, d)$, define its diameter to be $$ \text{sup} \{d(x, y) |x, y \in X\}. $$ If $X$ is compact, then its diameter is finite and there exist two points $x_0, y_0$ whose distance is equal to the diameter, because the function $$ d: X \times X \to \mathbb{R} $$ has compact, and hence bounded image.

The diameter of $f(X)$ is less than that of $X$, so they cannot be the same as each other. (Proof: say $d(f(x), f(y)) \geq \text{diameter}(X)$; then $d(x, y) > \text{diameter}(X)$ which is nonsense.)

hunter
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    (Note that it is not enough that $X$ be bounded for this to work -- it really has to be compact. For $(0, 1)$ you can take $x \mapsto x^2$, which decreases the distance between any two points but never decreases the diameter of the space, which is $1$. The point is that the diameter of a non-compact space need not be physically realized by two points, but is just a sup.) – hunter Dec 20 '20 at 02:42
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By way of contradiction, suppose $f$ is surjective.

Then $\forall x,y \in X$, we have $x_0,y_0 \in X$ such that $f(x_0)=x$ and $f(y_0) = y$. Then

$$d(x,y) = d(f(x_0),f(y_0)) < d(x_0,y_0)$$

Define the constant $M$ by

$$M := \sup_{(x,y) \in X \times X} d(x,y)$$

This exists since, trivially, the set of distances is nonempty. However, since metrics are continuous function and $M$ is compact, $d$ attains its supremum, and thus we may claim

$$M = \max_{(x,y) \in X \times X} d(x,y)$$

Thus, there exist some $x^*,y^*$ such that $d(x^*,y^*) = M$. However, there always exists a pair $(x_0,y_0)$ for any $(x,y)$ such that $d(x,y) < d(x_0,y_0)$. A contradiction is reached, and $f$ cannot be surjective.

Perhaps an example would be enlightening.

An easy one would be to take $X = [0,1]$ with the usual Euclidean metric $d$, and the function $f(x) = x/2$. Clearly meets the contraction criteria, but, for instance, is there $x$ such that $f(x)=1$?

I say contraction criteria because one may define a mapping $f$ to be a contraction on a metric space if $d(f(x),f(y)) \le k \cdot d(x,y)$ for every $x,y$ and for $k \in (0,1)$. Your function meets this criteria. Notice how, intuitively, the function $f$ "shrinks" the space, and reduces the distances between points.

PrincessEev
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