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Question:

Let $T = \{U \subseteq \mathbb{R} : U = \emptyset \text{ or } U = \mathbb{R}\text{ or } U = (−\infty, a) \text{ for some } a \in\mathbb{R}\}$.

Prove that $T$ is a topology on $\mathbb{R}$.

I know the axioms are:

  1. The empty set and $X$ are in $T$.
  2. The intersection of any finite collection of subsets of $X$ in $T$ is also in $T$.
  3. The union of any collection of sets in $T$ is also in $T$.

and can prove the 1st axiom easily but I am struggling to understand how to go about actually showing that the 2nd and 3rd axioms of topologies hold for any example question I attempt, not just this one.

I would appreciate if someone could give me the basics of the process involved in proving axiom 2 and 3 hold.

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You do know that the axioms of topology can change their order between one place and another, right? –  Asaf Karagila Apr 22 '12 at 15:59
    
Woops, I've edited the question so it makes a bit more sense now. –  Nicky Apr 22 '12 at 16:04
    
Your formulation of the third axiom is false. We require the union of any collection of sets in $T$ to be in $T$. –  Asaf Karagila Apr 22 '12 at 16:05
    
Indeed, Axiom 3 is not an axiom for topological spaces. –  Siminore Apr 22 '12 at 16:08
    
Edited again to correct 3! –  Nicky Apr 22 '12 at 16:13
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3 Answers

up vote 3 down vote accepted

What are your axioms 2 and 3?

You need to show:

$\ \ \ $1) $X$ and $\emptyset$ are in $T$.

$\ \ \ $2) Any union of elements of $T$ belongs to $T$.

$\ \ \ $3) Any finite intersection of elements of $T$ belongs to $T$


1) holds by the definition of $T$.


For 2):

Let $ \{ U_\alpha\mid \alpha\in I\}$ be a non-empty collection of elements of $T$. You need to verify that $O=\bigcup\limits_{\alpha\in I} U_\alpha\in T$. This is perhaps best done by considering cases.

If one of the $U_\alpha$ is $\Bbb R$, so is the union, and then $O\in T$.

If all $U_\alpha=\emptyset$, then $O=\emptyset\in T$.

Otherwise, let $\beta=\sup\{\alpha\mid\alpha\in I\}$. If $\beta=\infty$, you can show $O=\Bbb R\in T$. If $\beta$ is finite, you can show that $O=(-\infty,\beta)\in T$.


3) is easier to verify:

For the nontrivial case where no member of the finite collection of sets is empty, given a finite collection $\{(-\infty,\alpha_1),\ldots, (-\infty,\alpha_k)\}$ of elements in $T$ (here $a_k$ is allowed to be infinity), the intersection is $(-\infty, \min\{\alpha_1,\ldots,\alpha_k \})\in T$.

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Thanks! This helped a lot. –  Nicky Apr 22 '12 at 16:54
    
For the second axiom you wrote that $O$ is the collection and we want $O\in T$, but we actually want $\bigcup O\in T$. –  Asaf Karagila Apr 22 '12 at 18:02
    
@AsafKaragila eek! Thanks. –  David Mitra Apr 22 '12 at 18:10
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Hint: Use the fact that $\mathbb R$ is linearly ordered and order complete to deduce that:

  1. $(-\infty,a)\cap(-\infty,b)=(-\infty,\min\{a,b\})$, and
  2. that the union of intervals $(-\infty,a_i)$ for $i\in I$ is $(-\infty,\sup\{a_i\mid i\in I\})$.
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I guess that Axiom 2 is for you the fact the any union of open sets is an open set, while Axiom 3 is the same for the intersection of any two open sets. Well, in general there is no recipe. In your example, let's take the intersection of two open sets. If one of them is $\emptyset$, the intersection is empty. If one of them is $\mathbb{R}$, the intersection is the other open set. If both have the form $(-\infty,a)$ and $(-\infty,b)$, then their intersection is $(-\infty,\min\{a,b\})$. In any case, the Axiom holds true. The proof for an arbitrary union is slightly more involved, but not impossible.

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You guessed wrong! Axiom 2 is actually intersection!! :-) –  Asaf Karagila Apr 22 '12 at 16:09
    
For arbitrary unions, the empty set has no effect on the union and $\mathbb{R}$ in the union makes the union all of $\mathbb{R}$. So you only need to consider unions of sets of the form $(-\infty, a)$. If there is a common upper bound for a collection of sets of that form, then the union equals $(-\infty, a)$, where $a$ is the supremum. If there is a no common upper bound for the collection of sets, then the union is $\mathbb{R}$. –  Mikko Korhonen Apr 22 '12 at 16:13
    
Of course, for arbitrary unions you have to play with least upper bounds. This is the topology of half-lines on $\mathbb{R}$. –  Siminore Apr 22 '12 at 16:16
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