# Prove that $U$ is a topology on $\Bbb{R}$

I was going through some exercises about topology, and I got stock on one of them.

If $$X=\mathbb{R}$$ and $$u \in U$$ if and only if $$u$$ is a subset of $$\mathbb{R}$$ and $$\forall s\in u \exists t>s$$ such that $$[s,t) \subseteq u$$, where $$[s,t)=\{x \in \mathbb{R};s\leq x.

Now the book says that I need to proof that $$U$$ is a topology of $$X$$, but I don't know what to do with this one. Do I have to prove that $$[s,t)$$ is a topology? Is that the definition of a half open topology? Can I use it?

• A first course in algebraic topology by C. Kosniowski – AlejandroL.g Apr 16 '20 at 1:21
• To show that $U$ is a topology on $X$, you will need to show four things: (1) $\varnothing\in U$, (2) $X\in U$, (3) for every $A,B\in U$, $A\cap B\in U$, and (4) for every subset $S\subseteq U$, the union $\bigcup S\in U$. Of these four, (1) and (2) will probably be the easiest to show, and (4) will probably be the hardest. – user729424 Apr 16 '20 at 1:25
• For (4) it's helpful to keep in mind that $x\in\bigcup S$ iff $x\in A$ for some $A\in S$. – user729424 Apr 16 '20 at 1:30
• Also, it might help to use slightly different notation, and call it $\mathcal{U}$ instead of $U$. That way you can use lower-case letters such as $a,b,c,\ldots$ for elements of $\Bbb{R}$, uppercase letters such as $A,B,C,\ldots$ for subsets of $\Bbb{R}$, and uppercase script letters such as $\mathcal{A},\mathcal{B},\mathcal{C},\ldots$ for sets of subsets of $\Bbb{R}$. With this notation we'd use $\mathcal{U}$ instead of $U$ since $\mathcal{U}$ is a set of subsets of $\Bbb{R}$. – user729424 Apr 16 '20 at 1:35
• With this new notation, the properties you have to show are (1) $\varnothing\in\mathcal{U}$, (2) $X\in\mathcal{U}$, (3) for any $A,B\in\mathcal{U}$, $A\cap B\in\mathcal{U}$, and (4) for any subset $\mathcal{S}\subseteq\mathcal{U}$, the union $\bigcup\mathcal{S}\in\mathcal{U}$. – user729424 Apr 16 '20 at 1:37

So $$\mathcal{T}$$ is defined by

$$\mathcal{T}=\{O \subseteq \Bbb R: \forall s \in O: \exists t> s: [s,t) \subseteq O\}$$

We can check directly that this defines a topology on $$\Bbb R$$:

For $$O=\emptyset$$, the condition is satisfied voidly (a universal statement over the empty set is always true), so $$\emptyset \in \mathcal{T}$$.

For $$O=\Bbb R$$ we can just take $$t=s+1$$ for any $$s \in O$$ and the condition holds, so $$\Bbb R \in \mathcal{T}$$.

If $$O_1, O_2 \in \mathcal{T}$$ then let $$s \in O_1 \cap O_2$$ be arbitrary. Then $$s \in O_1$$, so the condition on open sets gives us $$t_1 > s$$ such that $$[s,t_1) \subseteq O_1$$. Also, $$s \in O_2$$ so the condition on open sets gives us $$t_2 > s$$ such that $$[s,t_1) \subseteq O_2$$. But then $$t=\min(t_1,t_2)>s$$ too and $$[s,t) \subseteq [s,t_1) \cap [s,t_2) \subseteq O_1 \cap O_2$$, and as $$s$$ was arbitrary, $$O_1 \cap O_2 \in \mathcal{T}$$, by definition. This shows that we have closedness under finite intersections.

If $$O_i, i \in I$$ is a family of sets from $$\mathcal{T}$$, then let $$s \in \bigcup_{i \in I} O_i$$ be arbitrary. Then for some $$j \in I$$, $$s \in O_j$$ by the definition of a union. Because $$O_j \in \mathcal{T}$$ we have $$t > s$$ such that $$[s,t) \subseteq O_j \subseteq \bigcup_{i \in I} O_i$$ and so $$\bigcup_{i \in I} O_i \in \mathcal{T}$$, and we have closedness under arbitrary unions.

We could also have shown that $$\mathcal{B}=\{[s,t): s, t \in \Bbb R: s < t\}$$ fulfills the axioms for a base for a topology (easy as the union is $$\Bbb R$$ and $$\mathcal{B}$$ is closed under finite intersections; facts that were implicitly used in the above proof). The resulting topology is called the Sorgenfrey line or (sometimes) the lower limit topology on $$\Bbb R$$.

• Thanks, really good answer, so we could say that $\mathcal{B}$ gives rise to a topology space? – AlejandroL.g Apr 16 '20 at 23:58
• @JoshuaLópezAraiza we can say that $\mathcal{B}$ is a base for some topological space, namely the one we just checked. – Henno Brandsma Apr 17 '20 at 9:00