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9:42 AM
in Mathematics, 33 mins ago, by shintuku
so I'm trying to prove the set $\Omega := \{\varnothing\}\cup\{[0,\infty) \in \mathbb{R}\}\cup\{(a,\infty):a\in\mathbb{R}_{0+}\}$ is a topology on $X=[0, \infty)$, and I'm doing the condition $\forall \Omega_a, \Omega_b$ in $\Omega$, we have $\Omega_a \cup \Omega_b \subset \Omega$.
 
Martin comes and then disappears
 
 
1 hour later…
11:00 AM
Let us denote $b=\inf I$, where $I\subseteq [0,\infty)$. Then we have
$$(b,\infty)=\bigcup_{a\in I}(a,\infty).$$
Now I have noticed that you have posted the question on the main site, so I guess you might get a more detailed response there:
0
Q: Verify that an arbitrary union of members of $\Omega$ belong to $\Omega$

shintukuLet $X$ be the ray $[0, \infty) \subset \mathbb{R}$, and let $\Omega$ consist of $\varnothing$, $X$, and all rays $(a, \infty) \subset \mathbb{R}_{0+}$. I'm trying to prove $\Omega$ is a topology on $X$, but I'm stuck at the idea of an arbitrary subcollection of sets for the proof that an arbitra...

 
woah, thanks a lot for taking up the question! @MartinSleziak
yeah, I've gotten stuck pretty hard with this question
I'll be processing what you've just said hehe
 
I would say that if you can show that $\bigcup_{a\in I}(a,\infty)\in\Omega$ for every $I$, the proof would be almost done.
 
right, that makes sense
 

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