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We have a topological space $X$ with bases $\mathcal{B_1}$ and $\mathcal{B_2}$. For $x\in X$, there is a $B_{1,x}\in\mathcal{B_1}$ and $B_{2,x}\in\mathcal{B_2}$ which contain $x$, by definition of basis.

Since it is a basis, without loss of generality $\mathcal{B_1}$ generates the topology $X$, and so for any open set $U$ in $X$ and $x\in U$ there is a basis element $B_{1,x}$ with $x\in B_{1,x}$ and $B_{1,x}\subseteq U$. Then it follows that such a basis element in $\mathcal{B_2}$, say $B_{2,x}$ contains such $B_{1,x}$, since $B_{2,x}$ is an open set; and likewise for $B_{1,x}$, which is open as well, so that there is a basis element $B'_{2,x}\subseteq B_{1,x}$, etc.

In this fashion, having fixed two bases of $X$, each point of $x$ "induces" a nested collection (or sequence, if it is countable) of 'alternating' basis elements, which may possibly 'terminate' if at some point along this nesting it holds that $B^\alpha_{1,x}=B^\beta_{2,x}$, where $\alpha,\beta$ are in some index sets which index the basis elements of $\mathcal{B_1}$ and $\mathcal{B_2}$, respectively, which participate in this nested collection.

Is this above correct?

Edit This may be a related question: Constructing a local nested base at a point in a first-countable space

A__A__0
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  • Of course the sequence you get is far from unique, and there's no guarantee that the basis elements in the sequence are going to get small. – bof Nov 10 '17 at 03:44
  • I suppose it is not unique in the sense that each basis element in the collection may contain a number of basis elements, each of which could be used to continue the nesting, so that at any point in the collection/sequence, we could 'branch off' onto another sequence; would you agree with that description? Regarding the basis elements getting small, you mean because the containment is not necessarily proper, right? Surely if the containment had to be proper, they would get 'small', or at least smaller – A__A__0 Nov 10 '17 at 03:47
  • Let both bases be the regular rational endpoint base for R. For x = 0, your sequence could progress (-1\n, 1/n), n= 1,2,3... .which would not terminate or if it did at the omega_0 th step would be the not open {0}. – William Elliot Nov 10 '17 at 03:56
  • @WilliamElliot Indeed, I agree – A__A__0 Nov 10 '17 at 04:06

1 Answers1

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You can try such a recursive construction, but it might very well fail at limit steps:

Suppose we have $x \in X$ and $B^1_0 \in \mathcal{B}_1$ that contains $x$. Then there is a $B^2_0 \in \mathcal{B}_2$ such that $x \in B^2_0 \subseteq B^1_0$, then we pick $B^1_1 \in \mathcal{B}_1$ such that $x \in B^1_1 \subseteq B^2_0$, etc.

So we try to construct $B^i_\alpha$, where $i=1,2$ and $\alpha$ an ordinal such that $\forall \alpha < \beta$: $x \in B^1_\beta \subseteq B^2_\alpha \subseteq B^1_\alpha$, and we can go on for $\alpha \in \omega$. and it can stabilise when we have some common $B \in \mathcal{B}_1 \cap \mathcal{B}_2$, containing $x$, that we can keep on choosing. In a regular space we could choose the closure of the base element to sit inside the prviosu one, and we stabilise at clopen common base elements, e.g. (we could have isolated points).

But at limit stages it's unclear what to do, we can only continue if $\cap_{\alpha<\beta} B^i_\alpha$ would have non-empty interior. But if a point is a $G_\delta$ we can easily get stuck at stage $\omega$ of a recursion, because then that intersection could be just $\{x\}$ very easily.

So you can make mutually nested decreasing base sequences, but even these might trivialise if you don't take precautions.

What do you want to achieve with these constuction ideas?

Henno Brandsma
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  • Thanks, my purpose is that I want to show that if a space has a countable basis, every basis contains a countable basis. My understanding is that these observations about containment will allow me to "bind" any basis to an existing countable one in a way that ensures that the arbitrary basis covers X countably.However it's a homework assignment so I don't want to ask about it directly. – A__A__0 Nov 12 '17 at 16:30
  • @A__A__0 I wrote a proof of the targeted fact as well. Search for it. It’s on this site. – Henno Brandsma Nov 12 '17 at 17:29