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I have to solve this question:

Given the following topology T on the real line R: T is generated by the set D = {ø, R, {[a,b)|a < b, where a and b are elements of R}}. Show that every subset [a,b) is also closed in the topological space (R,T).

We know that every subset [a,b) is open in (R,T) by definition (a set is open in a topology if it's a subset of that topology). To show that every subset [a,b) is also closed in the topology, I tried to show that the complement of [a,b), R\[a,b) = (-inf,a) U [b,inf), is open in the topology (by showing it's contained in the topology). Thus far I haven't succeeded!

Does anyone know how to solve this? Thanks in advance!

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You were doing good. Just consider that

$(-\infty,a)=\bigcup_{n\in\mathbb{N}}[-n-1+a,a)$ and hence it is open

and similarly

$[b,\infty)=\bigcup_{n\in\mathbb{N}}[b,b+n+1)$ and hence it is open.

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  • $\begingroup$ Uhm I don't fully understand this solution. I know that we can write (-inf, a) as ⋃n∈ℕ(−n,a), but why are you allowed to close the left side of the interval by substracting 1 and adding a? Regarding [b, ing); why would you add b and 1? Doesn't [b, inf) equal ⋃n∈ℕ[b,n)? Thanks for you reply! $\endgroup$ – titusAdam Feb 20 '17 at 12:12
  • $\begingroup$ Concerning the first point. The idea is that $a-1<a$ and hence I can "close". Concerning the second point. Since $n$ can be zero, adding $+1$ is to avoid the case $[b,b)$. Hope this helps! $\endgroup$ – Maczinga Feb 20 '17 at 12:17

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