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I don't see anything in my topology references to show that 2 bases generate the same topology. Is there a criteria?

Guest_topology
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  • Show that you can generate each base from the other. – copper.hat Feb 25 '15 at 20:15
  • Would if suffice to check that for every $x \in X$, if $B_1$ is a basis element containing $x$ from one of the bases, there is a $B_2$ from the other base so that $x \in B_2 \subset B_1$, and the vice versa check? – Guest_topology Feb 25 '15 at 20:28
  • Yes, because you can state this as saying for all $x \in B_1$, there is some $B_2 \subset B_1$ from the other base that contains $x$. Hence $B_1$ is generated by the collection of $B_2$s satisfying this condition. – copper.hat Feb 25 '15 at 20:32

4 Answers4

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If $\mathcal B$, $\mathcal B'$ are bases for topologies $\mathcal T$, $\mathcal T'$ on $X$, then $\mathcal T\subset \mathcal T'$ if and only if for each $B\in\mathcal B$ and $x\in B$, there is a $B'\in\mathcal B'$ such that $x\in B'\subset B$. To remember this, here is the analogy used in Munkres's Topology:

It may be easier to remember if you recall the analogy between a topological space and a truckload of gravel. Think of the pebbles as the basis elements of the topology; after the pebbles are smashed to dust, the dust particles are the basis elements of the new topology. The new topology is finer than the new one, and each dust particle was contained inside a pebble, as the criterion states.

So, $\mathcal T=\mathcal T'$ if and only if this condition is met for both bases.

Tarc
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Math1000
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  • Do you mean for each $x \in \mathcal{T}$? – Daniel Xiang Mar 16 '17 at 01:17
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    @Math1000, AS Munkres Stated, where the inclusion goes is quiet confusing, If "each dust particle was contained inside a pebble" then if $\tau \subset \tau'$ that is $\tau ' $ is finer than $\tau $ shouldn't the inclusion be such that if $U$ is an open set in $\tau'$ then $U \in \tau$. Why is it the other way around as in Lemma 13.3 in the book? – Cnine Feb 19 '18 at 19:24
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Suppose the bases are $\mathcal{B_1}$ and $\mathcal{B_2}$, generating $\mathcal{T_1}$ resp. $\mathcal{T_2}$.

Then $$\forall B_1 \in \mathcal{B_1} \forall x \in B_1 \exists B_2 \in \mathcal{B_2}: x \in B_2 \subseteq B_1$$ is equivalent to $\mathcal{T_1} \subseteq \mathcal{T_2}$.

So for equality we need this condition and the reverse $$\forall B_2 \in \mathcal{B_2} \forall x \in B_2 \exists B_1 \in \mathcal{B_1}: x \in B_1 \subseteq B_2$$ as well.

Henno Brandsma
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Suppose you have to bases $\mathcal{B}$ and $\mathcal{B}'$ which generate $\mathcal{T}$ and $\mathcal{T}'$ respectively. We want to show that $\mathcal{T}=\mathcal{T}'$. Notice that if $\mathcal{B}\subseteq\mathcal{T}'$, then $\mathcal{T}\subseteq\mathcal{T}'$ (the converse is similar). Thus it suffices to show that each set in $\mathcal{B}$ can be generated from sets in $\mathcal{B}'$ and vice versa.

rnrstopstraffic
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I'm guessing you have an explicit example you are trying to prove this result about? So you have two bases, $\mathcal{B} = \{B_\alpha\}_{\alpha \in \lambda}$ and $\mathcal{C} = \{C_\sigma\}_{\sigma \in \gamma}$. What you want to show is given any $B_\alpha$, there exist a collection of open sets in $\mathcal{C}$ such that $$B_\alpha = \bigcup_{\sigma \in \gamma'}C_\sigma$$ and for any $C_\sigma$ there is a collection in $\mathcal{B}$ such that $$c_\sigma = \bigcup_{\alpha \in \lambda'}B_\alpha$$ where $\gamma', \lambda'$ are just indexing sets, $\gamma' \subset \gamma$, $\lambda' \subset \lambda$. Hence for any open set in the topology generated by one basis, you know you can use the other basis to make the same open set.

graydad
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