Let [itex]\rho[/itex] be the uniform metric on [itex]\mathbb{R}^\omega[/itex](adsbygoogle = window.adsbygoogle || []).push({});

For reference, for two points:

[tex]x = (x_i)[/tex] and [tex]y = (y_i)[/tex] in [tex]\mathbb{R}^\omega[/tex]

[tex]\rho(x,y) = \sup_i\{ \min\{|x_i - y_i|, 1\}\}[/tex]

Now, define:

[tex]U(x,\epsilon) = \prod_i{(x_i - \epsilon, x_i + \epsilon)} \subset \mathbb{R}^\omega[/tex]

I need to show 2 things:

1) For [itex]0 < \epsilon < 1[/itex], the ball of radius [itex]\epsilon[/itex] about [itex]x[/itex] in the uniform metric, [itex]B_\rho(x,\epsilon) \neq U(x,\epsilon)[/itex]

2)[tex]B_\rho(x,\epsilon) = \bigcup_{\delta < \epsilon}{U(x,\delta)}[/tex]

-----

Now, first off, I can't understand how 1 can be true. It makes no sense to me. Am I misunderstanding the metric? To further obfuscate the issue, 2 seems to contradict 1. Afterall, doesn't:

[tex]\bigcup_{\delta < \epsilon}{\prod_i{(x_i-\delta, x_i+\delta)}}} = \prod_i{\bigcup_{\delta<\epsilon}{(x_i-\delta,x_i+\delta)}}}[/tex]

The union inside the product of the latter being [itex](x_i-\epsilon, x_i+\epsilon)[/itex]? Thus, isn't part 2 suggesting that, in fact, [itex]B_\rho(x,\epsilon) = U(x,\epsilon)[/itex]

Am I missing something big here? I feel like it's obvious the problem is incorrect, so I can't understand how the author ( Munkres' Topology First Edition, section 2-9 #6 for reference ) could make the mistake. Please, someone throw me a bone here, I'm juststumped

**Physics Forums | Science Articles, Homework Help, Discussion**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Uniform Metric vs Box Topology

**Physics Forums | Science Articles, Homework Help, Discussion**