Munkres Topology: Let $X$ be a metric space with metric $d$.

  • a) Show that $d:X\times X\to \mathbb{R}$ is continuous.
  • b) Let $X'$ denote a space with the same underlying set as $X$. Show that if $d:X'\times X' \to \mathbb{R}$ is continuous then the topology of $X'$ is finer than the topology of $X$.

This has been asked a lot previously, and the answers involve $B_d(x,r)$. Here are the other questions. I got really confused because Munkres does not use $d'$ for the latter $d$. So, I don't know which $d$ the $d$ in $B_d(x,r)$ refers.

If $X$ has a metric $d$ then the topology induced by $d$ is the smallest topology relative to which $d$ is continuous

How is the metric topology the coarsest to make the metric function continuous?

Two problems related to continuity of a metric from Munkres' topology book

Topology induced by metric space

Munkres Section 20 Exercise 3b. Proof verification.

There's an alternate proof that makes use of the identity function instead of $B_d(x,r)$ from mathstuffed's Collected solutions for Munkres (Topology)

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My questions:

  1. Is $d \circ i$ the $d$ in $(b)$ while $d$ is the $d$ in $(a)$? So the $d \circ i$ here is what I would call $d'$?

  2. I think both $d$ and $i$ are continuous and so $d \circ i$ continuous because both $d$ and $i$ are continuous. Is this correct?

  3. Why is $\mathcal T_{X' \times X'}$ finer than $\mathcal T_{X \times X}$? I think Exercise 18.3 would say that if $i$ is continuous then $\mathcal T_{X' \times X'}$ is finer than $\mathcal T_{X \times X}$. Is $i$ continuous? The proof sounds like $\mathcal T_{X' \times X'}$ is finer than $\mathcal T_{X \times X}$ because $d \circ i$, not $i$, is continuous. Did the proof somehow conclude $i$ is continuous because $d \circ i$ is continuous?

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    $\begingroup$ Whoa - posting two so similar questions in so quick succession; this gives me the impression you have only 24 hours to save the world by solving the problem. --- There is only one $d$. $\endgroup$ – Hagen von Eitzen Sep 23 '18 at 10:00
  • $\begingroup$ @HagenvonEitzen But $d$ has a different domain from $d \circ i$? $\endgroup$ – user198044 Sep 23 '18 at 10:41
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    $\begingroup$ There is not necessarily a metric $d'$ that generates the topology of $X'.$ Many topologies cannot be generated by metrics. $\endgroup$ – DanielWainfleet Sep 23 '18 at 11:26
  • $\begingroup$ @HagenvonEitzen I thought it is too much for one question. But I really hate that I can't understand this. I don't know if I'm over thinking or under thinking. $\endgroup$ – user198044 Sep 23 '18 at 11:34
  • $\begingroup$ @DanielWainfleet But $d'$ is $d \circ i$? $\endgroup$ – user198044 Sep 23 '18 at 11:35

You're overthinking the problem. We are given a space space $X$ with a topology $\mathcal{T}_X$ which is induced by a metric $d$, meaning that the open balls $B_d(x,r)$ are a base for $\mathcal{T}_X$.

Now we have $X'$ which is another topology on the same underlying set, with topology $\mathcal{T}_{X'}$. So on the space $X' \times X'$ we have the product topology of $\mathcal{T}_{X'}$ with itself, and it is given that the metric function $d$, which is also defined on $X'\times X'$ (it's the same set after all) is in fact also continuous on $X' \times X'$ in this product topology (both with the usual topology on the image $\mathbb{R}$). The question is to show that this implies that the unknown topology $\mathcal{T}_{X'}$ is finer than $\mathcal{T}_{X}$.

Now I think that your quoted solution goes wrong: it needs to show that $i$, the identiy map from $X' \times X'$ to $X \times X$ is continuous. Then the rest would go through. But the continuity of $i$ does not follow from the continuity of $d$ as far as I can see, and there is no argument given there. IMHO it tries to be too clever.

Another argument is simpler: Given that $d: X' \times X' \to \mathbb{R}$ is continuous, fix $x \in X'$. Then $d_x: X' \to \mathbb{R}$ given by $d_x(y) = d(x,y)$ is also continuous (it's just $d$ composed with an obvious map from $X'$ into $X' \times X'$) and then $B_d(x,r) = \{y : d(x,y) < r \} = (d_x)^{-1}[(-\infty,r)]$ is thus open in $X'$ too. This holds for all $x \in X'$ and all $r>0$, so the base of $\mathcal{T}_X$ (the $d$-open balls) are all open in $\mathcal{T}_{X'}$ and all open sets in $\mathcal{T}_{X}$ are unions of open balls so still open in $\mathcal{T}_{X'}$, hence $\mathcal{T}_{X} \subseteq \mathcal{T}_{X'}$.

The crux is to see the open balls, the base of the metric topology $\mathcal{T}_{X}$ as inverse images of open sets under the metric as a function.


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