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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0the subspace topology – 2012-03-06
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0@AsafKaragila: $d(x,0)
prevents $r$ from being that small. As far as I can see, the answer is that an open set $S$ must be conventionally open _and_ contain a punctured ball around the origin _and_ satisfy $tS\subseteq S$ for all $t\in (0,1)$. – 2012-03-06 -
0@Henning: Sleep deprivation... :-) – 2012-03-06
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0In other words, a set is open if it can be written $\{ xr \mid x\in S^1, 0
for some continuous $\varphi: S^1 \to (0,\infty]$. – 2012-03-06 -
0I was thinking whether it would be the subspace topology, like what yoyo suggested in the earlier comment, but I was unable to prove it. – 2012-03-06
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1It’s clearly not the subspaces topology: if $d(x,0)=2r>0$, then $B(x,r)$ is open in the subspace topology but not in the topology generated by this base. – 2012-03-06
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0I don't think the collection of balls is a basis. Small balls are relegated to regions nearer the origin, so just take two large balls far away from the origin that intersect in a very small set also far away from the origin. The intersection won't be open. However your collection is a sub-basis. By "generate" a topology do you mean we should close under finite intersection and then close under unions? Closing under unions isn't enough. – 2012-03-06
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0@Yeong-ChyuanChung: But to be a basis you need to be able to find a basis element around an arbitrary point in the intersection, not just the origin. – 2012-03-06
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0@JimConant: I see what you mean. If it is taken to be a subbasis, then what would the topology be? – 2012-03-06
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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