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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?

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  • $\begingroup$ $\beta$ is a chain already. That is the asumption. $\endgroup$ Commented Sep 11, 2019 at 16:43
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    $\begingroup$ 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. $\endgroup$ Commented Sep 11, 2019 at 16:55
  • $\begingroup$ You are right. I missed this part. $\endgroup$
    – Sanae
    Commented Sep 14, 2019 at 5:16

1 Answer 1

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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.

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  • $\begingroup$ 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 $L$? $\endgroup$
    – Sanae
    Commented Sep 14, 2019 at 5:35
  • $\begingroup$ 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 $L$? $\endgroup$
    – Sanae
    Commented Sep 14, 2019 at 5:42
  • $\begingroup$ @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) $\endgroup$ Commented Sep 16, 2019 at 21:58
  • $\begingroup$ Got it. Now it is more clear. $\endgroup$
    – Sanae
    Commented Sep 18, 2019 at 2:54

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