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Consider $\mathbb{R}^{2}\setminus\{0\}$. If I take the collection $\{ B(x,r)\setminus\{0\}: d(x,0)<r,r>0 \}$ as a subbasis, what is the topology generated?

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the subspace topology –  yoyo Mar 6 '12 at 2:26
    
@AsafKaragila: $d(x,0)<r$ 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 $t\in (0,1)$. –  Henning Makholm Mar 6 '12 at 2:26
    
@Henning: Sleep deprivation... :-) –  Asaf Karagila Mar 6 '12 at 2:27
    
In other words, a set is open if it can be written $\{ xr \mid x\in S^1, 0<r<\varphi(x)\}$ for some continuous $\varphi: S^1 \to (0,\infty]$. –  Henning Makholm Mar 6 '12 at 2:31
    
I was thinking whether it would be the subspace topology, like what yoyo suggested in the earlier comment, but I was unable to prove it. –  Yeong-Chyuan Chung Mar 6 '12 at 2:35

1 Answer 1

up vote 2 down vote accepted

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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And 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 $(0,x]\subseteq U$. –  Brian M. Scott Mar 6 '12 at 3:38

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