(Munkres, p. 126, Ex. 3)

Prove the following:

Let $X$ be a metric space with a metric $d$. Let $X'$ be a topological space that has the same underlying set as $X$; i.e., $X' =X$ but $X'$ might have a different topology on it. Suppose $d : X' \times X' \to \mathbb{R}$ is continuous. Then the topology of $X'$ is finer than that of $X$.


Outline: Since $d:X'\times X'\to\Bbb R$ is continuous, then for any $x\in X',$ the function $f:X'\to\Bbb R$ given by $f(y)=d(x,y)$ is continuous. It follows that all the open $d$-balls are open in the topology on $X'.$ Since the open $d$-balls form a basis for the topology on $X,$ then the topology on $X'$ is finer.

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    $\begingroup$ Thank you for your answer. If I may add a detail for other viewers, on your first sentence, it it quite helpful to note that projection mapping is an open map. Then your first sentence easily follows. $\endgroup$ – jachilles Sep 27 '13 at 18:31

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