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I have 100 sets $A_1,\ldots,A_{100}$. They are all subsets of $\Bbb R$. For each $A_i$ the complement of $A_i$ in $\Bbb R$ is a countable set.

$A= A_1 \cap A_2 \cap \ldots \cap A_{100}$.

$B$ is the complement of $A$. What is the cardinal number of $B$? There are 4 options as an answer:

  1. $0$
  2. finite number but not $0$
  3. $\aleph_0$
  4. c

I think that is can be $2$ or $3$ because we don't know if the 'complement of $A_i$ in $\Bbb R$ is a countable set' is all finite or one of them is $\aleph_0$.

Am I right?

The formal answer for this question is $3$.

TonyK
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user3523226
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    I think you're right. It would be easier for people to understand and answer the question if it the formulas were typeset with MathJax, however. See https://math.stackexchange.com/help/notation -- someone has already started to do this. – David K Jan 22 '19 at 13:39
  • What does A=A1 ^ A2 ^ ... ^ A100 mean? Is this supposed to be a tower of exponents, or something else? – saulspatz Jan 22 '19 at 13:44
  • I mean ∩Ai , sorry – user3523226 Jan 22 '19 at 13:50
  • The complement of the intersection is the same as the union of the complements, so you have the union of 100 countable sets. Typically, this could be 0, finite, or countable infinite. However, in some literature countable is used to refers only to infinite sets, in which case it's 3. – Todor Markov Jan 22 '19 at 13:55
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    If you take $A_i = \mathbb{R}$ for all $i$, then $A = \mathbb{R} $, each complement is empty (so countable) , and $B$ is empty with cardinality $0$. – JuliusL33t Jan 22 '19 at 13:55

2 Answers2

1

The answer depends only on what your definition of "countable" is. As you point out, if a countable set can be finite, then option $2$ is indeed a possibility. See this Wikipedia article: "Some authors use countable set to mean countably infinite alone."

Edited to add: As Keen-amateur points out, option $1$ is also possible if a countable set can be empty.

TonyK
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If you take countable to mean exactly countable, i.e. not finite, and not empty then $3$, is your only answer since

$$B = A^c = (\bigcap(A_i))^c = \bigcup(A_i)^c$$

which is a finite union of exactly countable sets. Otherwise, both $1$ and $2$ could be answers.

JuliusL33t
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