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(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 $X$.

Stefan Hamcke
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le4m
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2 Answers2

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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.

Cameron Buie
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    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. – le4m Sep 27 '13 at 18:31
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Here is the exact answer Cameron wrote, but I decided to write all of the steps. If $d:X'\times X'\rightarrow \mathbb{R} $ is continuous in the product topology, for every $(x_o,y_o)$ and fixed $\epsilon>0$, there is a basis element with $(x_o,y_o)\in U\times V$ such that

$$d(U\times V)\subset (d(x_o,y_o)-\epsilon, d(x_o,y_o)+\epsilon)$$

In particular $d(U\times y_o)\subset (d(x_o,y_o)-\epsilon, d(x_o,y_o)+\epsilon) $ and because $U$ is open in $X'$, we must have $f(x)=d(x,y_o)$ a continuous function with $f: X'\rightarrow \mathbb{R}$. Clearly, for every $r>0$:

$$f^{-1}(-\infty, r)=\{x'| d(x',y_o)<r\}=B(y_o,r)$$

So, open balls are open in $X'$ and hence all open elements is the usual metric topology are open in the topology of $X'$

Kadmos
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