Consider $\mathbb{R}^{2}\setminus\{0\}$. If I take the collection $\{ B(x,r)\setminus\{0\}: d(x,0)
Topology generated by a given subbasis
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0I don't think this space is any of the standard examples. Are there specific properties of this space that you are interested in? Also, if you take the topology on $\mathbb{R} \times \mathbb{R}$ generated by the subbase \{ B(x,r) : x \in \mathbb{R} \times \mathbb{R}, 0 < d(x,0) < r \}, then your space is clearly a subspace of this. Non-empty open sets in the $\mathbb{R} \times \mathbb{R}$ variant appear to be exactly the open star-convex sets about $0$ (open, of course, in the standard topology). – 2012-03-06
1 Answers
This is not a basis. Recall that a basis $\mathcal B$ must satisfy the property that for any $B_1,B_2\in\mathcal B$ and any $x\in B_1\cap B_2$, there is a third basis element $B_3$ such that $x\in B_3\subset B_1\cap B_2$. Now take $B_1=B((1,0),1.01)$ and $B_2=B((-1,0),1.01)$. Now $B_1\cap B_2$ is a narrow lens shape with two corners on the $y$-axis. Consider $x\in B_1\cap B_2$ such that $x$ is on the $y$ axis and very close to one of these corners. Then you can't possibly find a circle containing $x$ that also contains the origin. So there is no $B_3$.
One way to fix this is to consider the set you mentioned as a subbasis as opposed to a basis, where you take the generated topology to be arbitrary unions of finite intersections of subbasis elements.
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0And in that case it appears that the open sets are the deleted Euclidean nbhds $U$ of $0$ with the property that for each $x\in U$, the segment $(0,x]\subseteq U$. – 2012-03-06