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I'm not sure how to approach this problem. Let $A = \left\lbrace (-\infty,r),(r,\infty):r\in \mathbb{R}\right\rbrace$ and let $B = \left\lbrace (a,b) : a,b \in \mathbb{R}, a<b\right\rbrace$. Then the goal should be to show that $\tau(A) = \tau(B)$ where $\tau$ is the topology generated by A or B. I know the definition means I take every topology on $\mathbb{R}$ that contains $A$ and intersect all of them (likewise for B), and the natural way to prove this was to show containment both ways, but I'm not sure how to get started.

From reading my notes, it sounds like I'm supposed to show that I can express $(a,b)$ of the form $(-\infty, r)$ and $(r,\infty)$ using unions and finite intersections and vice versa, but I don't see how.

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  • $\begingroup$ Are the sets in $A$ of the form $(-\infty,r) \cup (r,\infty)$ or of the form $(-\infty,r)$ or $(r,\infty)$? $\endgroup$ – Bob Krueger Jan 13 '16 at 20:17
  • $\begingroup$ @Bob1123 That's how it's written in our assignment. I'm assuming it means "or" $\endgroup$ – Brenton Jan 13 '16 at 20:22
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Suppose $a < b$. Then $(a,b) = (a,\infty) \cap (-\infty,b)$. As both $(a,\infty)$ and $(-\infty,b)$ are in $A$, and topologies are closed under finite intersections, $(a,b) \in \tau(A)$ (minor argument required, if you go from definitions only). This holds for all $(a,b)$, so $B \subseteq \tau(A)$. As $\tau(A)$ is a topology that contains $B$ and $\tau(B)$ is the intersection of all such topologies, $\tau(B) \subseteq \tau(A)$.

On the other hand, for all $r$: $(r,\infty) = \cup_{t > r} (r, r+t)$ which is a union of elements of $B$ (namely open intervals), and as all topologies are closed under arbitrary unions, $(r,\infty) \in \tau(B)$. Similarly (check this) we have that all $(-\infty,r) \in \tau(B)$ as well. So $A \subseteq \tau(B)$ and the same argument as above gives $\tau(A) \subseteq \tau(B)$ and we have equality of topologies.

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Big Hint: What is $(-\infty, b) \cap (a,+\infty)$ ?

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  • $\begingroup$ Assuming $a<b$ that's just $(a,b)$. So it sounds like the argument is very easy that I can write any element of $B$ as an intersection of elements of $A$. So I just need to show that I can write any element of $A$ of the form of elements of $B$ as well? $\endgroup$ – Brenton Jan 13 '16 at 20:44
  • $\begingroup$ And if $b\le a$ then the intersection is empty, which also is open. Re how to conclude: you already know (yes?) that elements of $A$ are open in the usual topology, so they don't give rise to any new open sets. But just to be thorough, $(r, +\infty) = \bigcup_{s>r}(r,s)$ and similarly for the left "rays". $\endgroup$ – BrianO Jan 13 '16 at 20:51

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