# Alternate proof of if $d:X'\times X' \to \mathbb{R}$ is continuous then the topology of $X'$ is finer than the topology of $X$.

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)

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?

• 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 $d$. – Hagen von Eitzen Sep 23 '18 at 10:00
• @HagenvonEitzen But $d$ has a different domain from $d \circ i$? – user198044 Sep 23 '18 at 10:41
• There is not necessarily a metric $d'$ that generates the topology of $X'.$ Many topologies cannot be generated by metrics. – DanielWainfleet Sep 23 '18 at 11:26
• @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. – user198044 Sep 23 '18 at 11:34
• @DanielWainfleet But $d'$ is $d \circ i$? – 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.