Relationship between seminormed, normed, spaces and Kolmogrov top. spaces

by Ocifer
Tags: kolmogrov, normed, relationship, seminormed, spaces
 P: 30 I am having trouble with a result in my text left as an exercise. Let (X, τ) be a semi-normed topological space: norm(0) = 0 norm(a * x) = abs(a) * norm(x) norm( x + y) <= norm(x) + norm(y) My text states that X is a normed vector space if and only if X is Kolmogrov. It claims it to be trivial and leaves it as an exercise. I'm able to get one direction, but I'm not very happy with it. ------------------------------------------------------------ (=> direction) Assume X is normed, let x,y be in X, such that x != y. Then d(x,y) > 0. I am tempted to use an open ball argument, claiming that there exists an open ball about x which cannot contain y, but then how do I relate this notion of open balls in a metric space to the open set required by the Kolmogrov (distinguishable) condition? ------------------------------------------------------------------------------- In terms of the other direction, I am entirely lost (Kolmogrov and semi-normed imples normed). Can anyone provide some insight, I just took a course in Real Analysis last semester and I did quite well, and I'm having trouble generalizing my insights now to topological spaces. With the semi-normed space we lose discernability, and I have a hunch that the Kolmogrov condition patches that problem, but I just can't get there.
 P: 418 For the => direction you are on the right track. You want an open ball B with center x, but with radius small enough that y is not in B. d(x,y) seems to be the only number you have to work with so try choosing a radius based on that. If you can find a radius such that y is not in B, then you are done because open balls are open sets. In the other direction you wish to show $\|x\| = 0$ implies x=0. Suppose $x \not= 0$, then by the Kolmogorov condition you can find an open neighborhood U of either x or 0 which does not contain the other. Try to use this to find a radius r such that B(x,r) or B(0,r) is contained in U, in which case you can show $\|x\|\not=0$. EDIT: Here B(p,r) means the open ball with center p and radius r.