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I am in a introductory topology class and our class was given a problem to think about and I am having trouble approaching it. He gave us a set X = $\mathbb R^n$ and a base:

$\beta$ = {all open balls $\ B_{\delta,n}(x) | \forall x\in \Bbb R^n, \delta > 0 $}. The question is to show that this is in fact a basis. It is pretty trivial to show that given any value in $\mathbb R^n$ we can form an open ball in the space. I am not sure how to approach the problem with two open balls intersecting and indicating that we can place any open ball within the intersection. The professor gave a hint about using the triangle inequality but I am otherwise kind of stuck. This is not a homework question just a practice question to try. Thanks for any help!

PS: Sorry about the weird formatting.

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Hint: Suppose two open balls $B_{\delta,n}(x)$ and $B_{\delta',n}(y)$ intersect. Then we have some point $z\in B_{\delta,n}(x)\cap B_{\delta',n}(y)$, so $d(z,x)<\delta$ and $d(z,y)<\delta'$. Let $\epsilon>0$ be smaller than $d(z,x)-\delta$ and $d(z,y)-\delta'$. Can you show that $B_{\epsilon,n}(z)\subset B_{\delta,n}(x)\cap B_{\delta',n}(y)$?

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Let $y \in B_{\delta,n}(x)$. Then $\delta - d(y, x) > 0$. Choose $\epsilon$ such that $0 < \epsilon < \delta - d(y, x)$. Suppose $d(z, y) < \epsilon$. Then $d(z, x) \le d(z, y) + d(y, x) < \epsilon + d(y, x) < \delta$. Hence $B_{\epsilon,n}(y) \subset B_{\delta,n}(x)$. The assertion follows immediately from this.

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Thank you very much... I think this will help greatly! –  InsigMath Sep 21 '12 at 3:27
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