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I need to prove that the Sierpiński space, $\mathcal{\tau} = \{\emptyset, \{1\}, \{0, 1\}\}$, is a topology.

I have only just started on toplogy, and so far just know the basic axioms.

To prove that the union of any collection of subsets from T is in T, do I need to list every single possible union of subsets and check that it is in T?

i.e. $\{1\} \cup \{0,1\} = \{0,1\} \in \mathcal{\tau}$, $\emptyset ∪ \{0,1\} = \{0,1\} \in \mathcal{\tau}$ etc.

Or is there a much shorter and more concise method?

helpmepls
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    Note that ${\emptyset}\cup{0,1}={\emptyset,0,1}\neq {0,1}$. There is a difference between $\emptyset$ and ${\emptyset}$, so you have a typo in what you wrote in your second to last line. – JMoravitz Apr 12 '21 at 17:17
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    There are three elements, so it's not too laborious to check every (unordered) pair of elements in the topology – Dan Rust Apr 12 '21 at 17:17
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    That said, I agree with Dan, it isn't time-consuming to check each here. It can be sped up though by noting that $\emptyset\subset {1}\subset{0,1}$ and that if $A\subset B$ then $A\cup B=B$. – JMoravitz Apr 12 '21 at 17:18
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    The typo you made was that it should have been $\emptyset \cup {0,1}={0,1}$... not that it should have been ${\emptyset}\cup {0,1}={\emptyset,0,1}$ – JMoravitz Apr 12 '21 at 17:19
  • Sorry I said pair, but of course you need to check arbitrary unions. For finite topological spaces that turns out to be equivalent to checking pairs, but that's technically something that needs to be proved. – Dan Rust Apr 12 '21 at 17:21
  • Thank you for both of your comments. I was thinking that {∅, {1}}, {∅, {0, 1}}, {{1}, {0, 1}} etc were all subsets so I would need to check all of the combinations; but clearly I was overthinking and there is much less work to be done. – helpmepls Apr 12 '21 at 17:22

2 Answers2

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The empty set adds nothing to a union, so we need not consider unions with the empty set.

Any union in which one of the sets is the universe $\{0, 1\}$ will just equal $\{0, 1\}$, so we can dispense with those.

Suddenly there's nothing left to check.

Austin Mohr
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    Similarly with intersections: if one is $\emptsyet$ so will the intersection be, and we know it’s in the topology already. And ${0,1}$ (or $X$) in the intersection does nothing.. – Henno Brandsma Apr 12 '21 at 21:24
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Checking every subset isn't too much work, but there is a way to do them simultaneously. You can observe that $T$ is totally ordered under the subset relation, i.e. $\emptyset \subseteq \{0\} \subseteq \{0,1\}$. Then for any subset $S \subseteq T$, the union of elements of $S$ is the maximal element of $S$ under inclusion, which is still in $T$. The same logic applies to intersections, where you instead get the minimal element.

How concise this argument really is depends on your standard of rigor. For instance, you'd have to prove that the union is the maximum and the intersection is the minimum. You may also want to specially consider the cases of the empty union and intersection. This isn't too hard, but it adds length. But regardless, if you were to do every union as a separate case, you'd end up using this same idea every time.

J. W. Tanner
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paul blart math cop
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