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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 $[s,t)=\{x \in \mathbb{R};s\leq x<t\}$.

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?

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    $\begingroup$ A first course in algebraic topology by C. Kosniowski $\endgroup$ – AlejandroL.g Apr 16 '20 at 1:21
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    $\begingroup$ 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. $\endgroup$ – user729424 Apr 16 '20 at 1:25
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    $\begingroup$ For (4) it's helpful to keep in mind that $x\in\bigcup S$ iff $x\in A$ for some $A\in S$. $\endgroup$ – user729424 Apr 16 '20 at 1:30
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    $\begingroup$ 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 $\Bbb{R}$. $\endgroup$ – user729424 Apr 16 '20 at 1:35
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    $\begingroup$ 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 $\bigcup\mathcal{S}\in\mathcal{U}$. $\endgroup$ – user729424 Apr 16 '20 at 1:37
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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 $\Bbb R$.

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

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