- #1
Rasalhague
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Kreiszig: Introductory Functional Analysis with Applications, example 1.3-9, states that the metric space [itex]l^{\infty}[/itex] is not seperable. He invites the us to consider the bijection between the interval [0,1] and the set Y of sequences, [itex]y = (\eta_1, \eta_2, ...)[/itex], whose terms consist of no number other than a 1 or a 0:
[tex]\hat{y}\in \left [ 0,1 \right ] = \sum_{k=1}^{\infty} \frac{\eta_k}{2^k}.[/tex]
The existence of this bijection, and the fact that [0,1] is uncountable, shows that Y is uncountable.
Consider the uncountably many balls of radius 1/3, each centered on one y in Y. (Whether open or closed is unstated, so I guess that's not relevant.) These balls don't intersect.
Therefore M is uncountable. M was chosen arbitrarily. Therefore, he concludes, [itex]l^{\infty}[/itex] has no dense subset which is countable. In other words, [itex]l^{\infty}[/itex] is not seperable.
My question. Am I right in thinking this assumes that no ball can contain a limit point of M which is not in M? If so, why is that?
[tex]\hat{y}\in \left [ 0,1 \right ] = \sum_{k=1}^{\infty} \frac{\eta_k}{2^k}.[/tex]
The existence of this bijection, and the fact that [0,1] is uncountable, shows that Y is uncountable.
Consider the uncountably many balls of radius 1/3, each centered on one y in Y. (Whether open or closed is unstated, so I guess that's not relevant.) These balls don't intersect.
If M is any dense set in in [itex]l^{\infty}[/itex], each of these nonintersecting balls must contain an element of M.
Therefore M is uncountable. M was chosen arbitrarily. Therefore, he concludes, [itex]l^{\infty}[/itex] has no dense subset which is countable. In other words, [itex]l^{\infty}[/itex] is not seperable.
My question. Am I right in thinking this assumes that no ball can contain a limit point of M which is not in M? If so, why is that?