# Question about Uniform Spaec with a nested space

This question is from the book "General Topology" written by John Kelly and it is Exercise D in Chapter 6, Page 204.

For definition of uniform space and the topological generated by the uniform, please refer to this wiki link

Used fact: (Metrirization Theorem, Thm 6.13) A uniform space is pseudo-metrizable iff its uniformity has a countable base

($$X, U$$) is a uniform space and $$\tau_{U}$$, the topology generated by the uniform is Hausdorff. Suppose $$\beta$$ is a base for $$U$$ and can be linearly ordered by inclusion. There are two conditions:

a. ($$X, U$$) is pseudo-metrizable.

b. Intersection of any countable family of open sets in ($$X, \tau_{U}$$) is open.

Show that either a) or b) holds for the uniform space ($$X, U$$).

The direction "$$a) \implies (\neg b)$$" is easy because a uniform space is Hausdorff (given the uniform topology) iff it is $$T_1$$. Assuming $$a)$$ is true, let {$$V_k$$|$$k \in \omega$$} be the base of uniform $$\beta$$ and then $$\cap_{k \in \omega}V_k[x]$$ = {$$x$$}, which is close. I have difficulty proving the other direction, especially in understanding how to use the linear ordering of $$\beta$$.

P.S.: Here is a possibly dumb question: In this question, under what circumstances can I turn $$\beta$$ into a chain?

• $\beta$ is a chain already. That is the asumption. Commented Sep 11, 2019 at 16:43
• As to your argument for $(a) \implies (\lnot b)$: this is invalid, as a closed set can be open too, so this is no counterargument. $X$ could be the discrete uniformity and then (a) and (b) both hold. Commented Sep 11, 2019 at 16:55
• You are right. I missed this part. Commented Sep 14, 2019 at 5:16

In the version of Kelley I have this is exercise D on p. 204 and is stated as

Let $$(X,\mathscr{U})$$ be a Hausdorff uniform space and suppose that a base $$\mathscr{B}$$ for $$\mathscr{U}$$ is linearly ordered by inclusion. The either $$(X,\mathscr{U})$$ is metrizable or the intersection of every countable family of open subsets of $$X$$ is open.

The proof is due to a simple observation:

Suppose $$(L, \le)$$ is a linear order. Then one of the two following statements must hold (where $$[L]^\omega$$ is the set of all countable subsets of $$L$$):

$$\exists A \in [L]^\omega: \forall x \in X :\exists a \in A: a \le x\tag{1}$$

$$\forall A \in [L]^\omega: \exists x \in X: \forall a \in A: x < a \tag{2}$$

These are clearly logical negations of each other so it's a simple case of $$\phi \lor \lnot \phi$$ having to hold for any $$\phi$$ (tertium non datur). $$(1)$$ says that $$L$$ has a countable downwards-cofinal subset, and $$(2)$$ that every countable subset has a lower bound in $$L$$.

We apply this to the linearly ordered set $$(\mathscr{B}, \subseteq)$$ and $$(1)$$ tells us that $$(X,\mathscr{U})$$ has a countable base (in the uniformity sense) and so $$(X,\mathscr{U})$$ is metrisable (it's already Hausdorff), and $$(2)$$ implies the fact about countable intersections of open subsets, as is easily seen.

• Thank you for your help, Henno. I have a few questions about your explanation. You mentioned $2$) implies that every countable subset has a lower bound in $L$. In this question, I believe $X$ will be replaced with $\mathscr{U}$ and could you explain how do you know that lower bound $x$ is in $L$? Commented Sep 14, 2019 at 5:35
• Here is a possibly dumb question again .... In your explanation, the order defined within $L$ is from the order defined within $X$. Even when $\le_{L}$ is different from $\le_{X}$, negations between $1$) and $2$) always exists. In this case do we really use the condition "$(L, \le) is linearly ordered$" or is this why you conclude the lower bound is in $L$? Commented Sep 14, 2019 at 5:42
• @SanaeKochiya The order $L$ is replaced by $\mathscr{B}$ which is linearly ordered by inclusion. There is no further requirement; the argument works in any linealry ordered set. (No order on $X$ is assumed or needed) Commented Sep 16, 2019 at 21:58
• Got it. Now it is more clear. Commented Sep 18, 2019 at 2:54