Re: Bounded sets in tvs

On 4 Juni, 13:19, David C. Ullrich <dullr...@xxxxxxxxxxx> wrote:
On Tue, 3 Jun 2008 23:35:51 -0700 (PDT), tttttes...@xxxxxxxxx wrote:
Here are two definitions of a bounded set A in a topological vector
space.

1. For all nbhds B of 0 there is an s > 0 such that A subset of tB for
every t > s.

2. For all nbhds B of 0 there is an s > 0 such that A subset of sB.

Why does 1 follow from 2?

Seems like it must have something to do with the continuity
of scalar multiplication in a TVS.

Let's see. Assume (2), and assume that B is a neighborhood of 0.
By continuity of scalar multiplication there exist a neighborhood
of zero B' and a number delta > 0 such that tx is in B for every
x in B' and every t with |t| < delta. By (2) there exists s > 0 such
that A is contained in sB'.

Now assume that t > s/delta and a is an element of A. Then
there exists b' in B' with x = sb'. Since s/t < delta it follows
that  b = (s/t)b' is in B, and x = tb. QED.

T

David C. Ullrich

Thank you.
.

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