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I am stuck on the following thing, which I do not know to be true or not.

If we have $X$ and $\mathcal{T_1}$ and $\mathcal{T_2}$ be topologies with neighborhood basis at each $x\in X$, $\mathcal{B}^x_1$ and $\mathcal{B}^x_2$ respectively, is it true that these topologies are equivalent if for every $B_1\in \mathcal{B}_1$ there exists a $B_2\in \mathcal{B}_2$ such that $B_2\subseteq B_1$, then the topologies must be same.

I am getting stuck because I cannot see how this implies, if it even does, however the elements of $\mathcal{B}^x_1$ and $\mathcal{B}^x_2$ are open in the other topology. I also guess I have the question of does neighborhood even have to have only open sets in it. As you can see I am generally confused, so any help would be appreciated.

Also if this is not true are there conditions that make it true?

Atticus Christensen
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    $\mathcal T_1\subseteq\mathcal T_2$ iff every neighborhood of $x$ with regards to $\mathcal T_1$ is also a neighborhood of $x$ with regards to $\mathcal T_2$, which is equivalent to the property you describe in the second paragraph. – Stefan Hamcke Mar 06 '13 at 00:32
  • I'm sorry. Why is that? – Atticus Christensen Mar 06 '13 at 00:37
  • do you mean the first or the second equivalence? – Stefan Hamcke Mar 06 '13 at 00:38
  • Actually. After some thinking I see why it is true, and I think I understand the answer to this question. Thank you, for your comment. – Atticus Christensen Mar 06 '13 at 00:40
  • By the way: If your interested how a topology can also be defined via neighborhood bases take a look at http://math.stackexchange.com/questions/321451/going-from-a-fundamental-system-of-neighborhoods-to-a-topology-and-vice-versa – Stefan Hamcke Mar 06 '13 at 00:44

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Ok, I've edited my answer.

Let $\mathcal{T}_1$ be the topology generated by $\mathcal{B}_1$, and $\mathcal{T}_2$ the topology generated by $\mathcal{B}_2$.

Suppose $U$ is open in $\mathcal{T}_1$. Then if $x\in{U}$ we know we can find a $B_1\in{\mathcal{B_1}^x}$ such that $x\in{B_1}\subset{U}$. By the property you mention, we can also find a $B_2\in{\mathcal{B}_{2}^x}$ such that ${B_2}\subset{B_1}$, and so $x\in{B_2}\subset{U}$. Therefore, $U$ is open in $\mathcal{T}_2$ also.

Does this make sense?

Axiom
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  • I guess my question was that could you phrase this differently if you were talking about the neighborhood bases. That is if $B_1$ is in the neighborhood basis for $x$ in $\mathcal{T_1}$, is it necessary to check that for all $y\in B_1$ there is a $B_2$ in the neighborhood basis for $x$ in $B_2$ such that $y\in B_2\subseteq B_1$ or is it necessary only to check this for $x=y$. – Atticus Christensen Mar 06 '13 at 00:19
  • Oh, I am sorry, I had completely forgotten that there were even something called "neighbourhood bases" (I assumed you talked about ordinary bases). By the way, what do you really mean by $\mathcal{B}1$? The collection of all neighbourhood bases at all points, or just at one particular point? Or do you use the notation $\mathcal{B}{1}(x)$ for the latter? – Axiom Mar 06 '13 at 00:32
  • one particular point. I will edit in some $x$s to show this – Atticus Christensen Mar 06 '13 at 00:33
  • Yes. Thank you! – Atticus Christensen Mar 06 '13 at 00:40