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Let $$\mathcal{B}_0=\{[a,b):a,b\in\mathbb{R},a<b\}$$ and $$\mathcal{B}_1=\{[a,b):a\in\mathbb{Q},b\in\mathbb{R},a<b\}.$$

Let $\sigma$ and $\tau$ be the topologies over $\mathbb{R}$ generated by the basis $\mathcal{B_0}$ and $\mathcal{B_1}$, respectively.

Are these spaces metrizable?

a) $(\mathbb{R},\sigma)\times(\mathbb{R},\sigma)$

b) $(\mathbb{R},\sigma)\times(\mathbb{R},\tau)$

c) $(\mathbb{R},\tau)\times(\mathbb{R},\tau)$

Well, for the part $a)$, $(\mathbb{R},\sigma)$ is just the Sorgenfrey line and it is easily shown that $(\mathbb{R},\sigma)\times(\mathbb{R},\sigma)$ is not normal by Jones' lemma. Hence this space cannot be metrizable because every metric space is normal.

How about b) and c)? Could you give me any hint?

I can use some metrizations theorems. For example Urysonh theorem (regular + second countable implies metrizable) and Nagata-Smirnov (though I think Nagata-Smirnov is a too heavy theorem).

Thank you all.

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For (b): $R_1=(R,\sigma)$, which is not metrizable, is homeomorphic to the subspace $R_1\times \{0\}$ of the space $R_1\times (R,\tau)$. A metrizable space cannot have a non-metrizable subspace......For (c), if $x\in Q$ let $B_x= \{[x,x+q):q\in Q^+\}.$ If $x\in R\backslash Q$ let $B_x=\{[q_1,q_2):q_1,q_2\in Q\land q_1<x<q_2\}.$ Then in $(R,\tau)$ the family $B_x$ is a local base at $x$ and every member of $B_x$ is closed. So $(R,\tau)$ is totally disconnected and is therefore a $T_{3\frac {1}{2}}$ space.From the Urysohn theorem it is metrizable. Hence its square is metrizable....... Remark : Another way to show that the Sorgenfry line is not metrizable: If ($X,d)$ is a metric space and $D$ is a dense subset of $X$ then $\{B_d(p,q) :p\in D\land q\in Q^+\}$ is a base. So a separable metrizable space is second-countable. Now $Q$ is dense in $ R_1=(R,\sigma)$ but if $B$ is any base for $R_1,$ then for $x\in R$ let $y_x>x$ and $f(x)\in B$ such that $[x,y_x)\subset f(x)\in B.$ Then $f:R\to B$ is 1-to-1 (because $\min f(x)=x$) so $B$ is uncountable.

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  • $\begingroup$ This post needs some "air" ... $\endgroup$
    – Déjà vu
    Jan 3, 2016 at 6:00
  • $\begingroup$ @ringe . What do you mean ? $\endgroup$ Jan 3, 2016 at 6:01
  • $\begingroup$ @user254665 I see, thank you. It seems that the same argument can be given for part a), so we don't have the need to use Jones' Lemma. Also, now that I think, for the part c) we would only need to think about if $(\mathbb{R},\tau)$ is metrizable or not. It's second countable (a basis is family of sets $[q,q+1/n),q\in\mathbb{Q}$). But could it be regular? $\endgroup$
    – JonSK
    Jan 3, 2016 at 6:16
  • $\begingroup$ $(R,\tau)$ is $T(3\frac {1}{2})$ so it is $T(3)$. And it's obviously Hausdorff. And $\{[q,q') :q,q'\in Q\land q<q'\}$ is a countable base $\endgroup$ Jan 3, 2016 at 6:39
  • $\begingroup$ I just realized that JonSK 's last comment, above, must have arrived before I added the answer to (c). $\endgroup$ Jan 3, 2016 at 7:08

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