Some topology on R is generated by a basis B, where B is the collection of the intervals [a,b) a,b in R, a < b. How to generate the whole R and phi?
Basis B is the collection of the intervals [a,b), how to generate R?
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real-analysis
general-topology
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0The more interesting question is: how to generate $\varnothing$? – 2017-01-29
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0See https://en.wikipedia.org/wiki/Lower_limit_topology – 2017-01-29
2 Answers
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You can do $\mathbb R=\cup_{n\in \mathbb Z} [n,n+1)$.
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Two ways. Suppose we have a base $\mathcal{B}$ for a topology, then we can describe the topology $\mathcal{T}$ generated by $\mathcal{B}$ as follows:
- The set of all unions of subcollections of $\mathcal{B}$, formally
$$\mathcal{T} = \{ \cup \mathcal{B'} : \mathcal{B'} \subseteq \mathcal{B}\}$$
here we can get $\emptyset$ for $\mathcal{B'} = \emptyset (\subseteq \mathcal{B})$ and $X = \cup \mathcal{B}$, which is one of the conditions for being a base for the topology. In our case we can of course also say in concreto that $$\mathbb{R} = \cup_{n \in \mathbb{Z}} [n,n+1)$$ or $$\mathbb{R} = \bigcup\{[x,x+1): x \in \mathbb{R} \}$$
- $O \in \mathcal{T}$ iff $\forall x\in O: \exists B_x \in \mathcal{B}: x \in B_x \subseteq O$
(which should be familar for metric spaces and $\mathcal{B}$ the collection of open balls).
In that case $\emptyset$ is open as there are no $x \in \emptyset$ to test (it's a void universal statement, which is true) and $X \in \mathcal{T}$ because we just take $x \in [x,x+1) \subseteq \mathbb{R}$ for any $x \in \mathbb{R}$.