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I am having a little difficulty understanding how to approach this question. When S = $2^R$ and T = $2^S$, would that mean that the elements of S = $\{2^1,\space 2^a\}$ and T = $\{2^2,\space 2^{2a}\}$?

By that understanding, would $|S| = 2$, $|T| = 2$, $|S \cup T| = 4$, and $|S \cap T| = \emptyset$?

I am completely new to these topics and I am slowly learning. Your patience is much appreciated. Thank you.

Asaf Karagila
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MH10
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1 Answers1

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I think the reference is to the power set.

So $S$ is the set of subsets. Here we get $S=\{\emptyset,\{1\},\{\alpha\},R\}$.

Now $|T|=2^4=16$.

Can you list the elements?

calc ll
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  • Could you elaborate further, Please? – MH10 Jul 08 '22 at 04:57
  • We're considering subsets here. That's the key. So in the case of $T$, it's actually the set of subsets of a set of subsets. I'll give some of the $16$ elements, and you try to get the rest. $T$ contains, for instance, ${{1}},{{\alpha}},{{1},R},{{\alpha}, R}$. So there's double set brackets. There are $12$ more elements. And always remember the empty set and the whole set, $S$. – calc ll Jul 08 '22 at 05:08
  • Oh, okay. I see. Would the elements be: [∅, {∅}, {{1}}, {{a}}, {R}, {∅, {1}}, {∅, {a}}, {∅, {R}}. {{1}, {a}}, {{1}, {R}}, {{a}, {R}}, {∅, {1}, {a}}, {∅, {1}, {R}}, {{a}, ∅, {R}}, S] ? – MH10 Jul 08 '22 at 06:35
  • @MikeHong, that is very close, but you are missing one three-element subsets of $S$. – Mees de Vries Jul 08 '22 at 06:49
  • Alright, sweet. I got it. Thank you all very much! – MH10 Jul 08 '22 at 08:56