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Let $(X,d)$ be a compact metric space, and $f \colon (X,d)\to \mathbb{R}$ is a continuous function such that if $x,y \in X$ and $x \neq y$ than $f(x) \neq f(y)$.

Let $t \colon X \times X \to \mathbb{R}$ where $t(x,y) = |f(x)-f(y)|$.

I want to prove that $t$ is a metric and that $t$ and $d$ are equivalent.

Proving that $t$ is a metric is easy. In order prove that these two are equivalent, I am thinking about showing that every subset of $X$ is $d$-closed if and only if it is $t$-closed, considering that every singleton set $\{x\}$ is closed in both $(X,d)$ and $(X,t)$. Is this a correct way of thinking ?

Jendrik Stelzner
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Stef M
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  • What do you mean by the metrics are equivalent? Just that if you know the properties of the function hold then $t$ is a metric and visa versa? – AnotherPerson Jan 06 '16 at 11:53
  • @El Chapo Two metrics, say $(X,t),(X,s)$ are considered to be equivalent when they generate the same topology on $X$ – Stef M Jan 06 '16 at 12:02
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    In my opinion it is more natural to show that two open balls with respect to the two metrics are contained in each other. That for a given $t$-ball you find a $d$-ball which is contained in the former is merely the definition of continuity. For the other way round you may use that $f$ is indeed uniformly continuous, bounded and the image is compact in $\mathbb{R}$ (and maybe also injectivity of $f$ helps). – M.U. Jan 06 '16 at 13:01

1 Answers1

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Your approach won't bring you very far. Singletons are closed in every metric space, so this doesn't help you much. Also, the compactness of $X$ is crucial for the proof, while your approach doesn't use it at all.

Try showing that every $t$-open set is also $d$-open and vice versa.

It is easy to show that every $t$-open set is $d$-open. This follows directly from the $\epsilon$-$\delta$-characterization of continuity and doesn't even need the compactness of $X$.

For the other direction, consider the inverse function $f^{-1}: f(X) \to X$. Since $(X, d)$ is compact, this function is continuous*. Now let $U \subset X$ be $d$-open and $x \in U$. Then there exists a $\delta > 0$ so that $U_\epsilon(x) \subset U$. Now choose an $\epsilon > 0$ so that $f^{-1}(U_\epsilon(f(x))) \subset U_\delta(f^{-1}(f(x))) = U_\delta(x)$ [this is the continuity of $f^{-1}$ at the point $f(x)$]. But this means $t(x, y) < \epsilon \implies d(x, y) < \delta$ (!), i.e. the $\epsilon$-neighborhood of $x$ with respect to $t$ is a subset of $U_\delta(x) \subset U$. Therefore we've shown that $U$ is $t$-open.

(*) To see this, observe that a function is continuous iff the preimage of every closed set is closed. If $M$ is closed, then it is also compact [since $X$ is compact] and therefore $(f^{-1})^{-1}(M) = f(M)$ is compact. Since $\mathbb{R}$ is Hausdorff, this implies the closedness of $f(M)$.

Dominik
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