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Both of the previous examples appear to be extendable to
Based on my limited understanding of Urysohn's lemma, the open sets need to be separable, and as you can see from the previous examples in this thread, the open sets are "concentric" (stretching the meaning of this term) and thus not separable.
The following variation of the previous example also seems to work:
X is the set of all (x,y,z) such that
Along the same lines as the first example, consider the set of all spheres of radius less than or equal to 1 (excluding the interior) and centered at (0,0,0), i.e.,
I think that I have found a simple proof. Would appreciate confirmation or corrections ...
Assume (Z,d) is a metric space that has T as its collection of open sets.
Assume d(1,2) = x and consider the open ball of radius x that is centered at 1, i.e., B(1;x).
B(1;x) must be one of the open sets in T and also contain 1 which implies that B(1;x) = O1. This implies that 2 is an element of B(1:x) since B(1;x) = O1. On the other hand, we assumed that d(1,2) = x which implies 2 is not in B(1;x) - a contradiction.
Would also appreciate any other simple examples like this, i.e., topological spaces that are not metrizable.
Concerning topological spaces that are not metrizable, I found the following example in Mendelson's Intro to Topology:
Let Z be the set of positive integers. For each positive integer n, let
On = {n, n+1, n+2, ...}.
Let T = { O1, O2, O3, ... }
The claim is that (Z, T) is a non-metrizable topological space.
Would appreciate any help in proving this (using basic results on metric spaces). Note the Mendelson's book does not get into the various metrization theorems at this point.
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