show that $\mathbb{R}^J$ in box topology is completely regular where $J$ is any set. I know it is completely regular in case uniform topology and box topology is finer than uniform topology. Therefore, there are more closed sets in box topology
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What is $RJ$? Do you mean $R^J$? – anomaly Jul 25 '16 at 21:26
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As its author notes at the end, a very minor adaptation of this proof that $\Bbb R^\omega$ with the box topology is completely regular will do the job. – Brian M. Scott Jul 25 '16 at 21:35
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Yes, I mean $R^J$ – Gob Jul 25 '16 at 22:11
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Does this answer your question? How to prove that $\mathbb R^\omega$ with the box topology is completely regular – PatrickR Apr 13 '23 at 23:30