Talk:Metric Induces Topology

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The following are contradictory:

$\tau_{\left({S, d}\right)} := \left\{{N_\epsilon \left({a}\right): \epsilon \in \R, a \in S, B_\epsilon \left({a}\right) \subseteq S}\right\}$

Let $\tau_{\left({S, d}\right)}$ be the set of all $X \subseteq S$ which are open in the sense that:

$\forall y \in X: \exists \epsilon \left({y}\right) > 0: B_{\epsilon \left({y}\right)} \left({y}\right) \subseteq X$

where $B_{\epsilon \left({y}\right)} \left({y}\right)$ is the open $\epsilon \left({y}\right)$-ball of $y$.

The latter does in fact form a topology; I think that $\tau$ should be topology corresponding to the basis $\left\{{B_\epsilon \left({a}\right): \epsilon \in \R, a \in S, B_\epsilon \left({a}\right) \subseteq S}\right\}$. This comes down to the second statement. --Lord_Farin 16:32, 18 January 2012 (EST)

Yes, I think you're right. I've checked back to my Steen and Seebach where they say (in their original notation to ensure accuracy):
"It is possible to use a metric to define a topology on $X$ by taking as a basis all open balls $B(x, \epsilon) = \{y \in X|d(x,y)<\epsilon\}$."
... which boils down to what I translated it into in the page itself. I believe I forgot to point out that $\{B(x, \epsilon)\}$ is in fact a basis.
I'm not well at the moment (brain starting to do unpleasant things, it's a function of extreme age, I believe) so feel free to amend this page as you think it ought to be, and then I'll take another look and see if it makes consistent sense with what this book says. --prime mover 17:06, 18 January 2012 (EST)


The definition should be separated from the proof, to conform to house style. --Lord_Farin 04:18, 24 February 2012 (EST)