Bachelier
Apr1-11, 03:11 AM
Not every separable complete bounded metr. space X is seq. compact. Correct?
I'm thinking of any subset in \mathbb{N} \ or \ \mathbb{R} with the discrete metric
Like consider (\mathbb{A}, d) \ such \ that \ \mathbb{A} \subset \mathbb{R} \ such \ as \ [0,n] \with \ n =1, ...., 100. \ and \ d \ is \ the \ discrete \ metric.
What do you think?
I'm thinking of any subset in \mathbb{N} \ or \ \mathbb{R} with the discrete metric
Like consider (\mathbb{A}, d) \ such \ that \ \mathbb{A} \subset \mathbb{R} \ such \ as \ [0,n] \with \ n =1, ...., 100. \ and \ d \ is \ the \ discrete \ metric.
What do you think?