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This is based off my last question How would the intersection of two uncountable sets be finite?

Here is the problem(from Discrete Mathematics and its Applications) enter image description here

The book's definition on countable enter image description here

And the definition of having the same cardinality enter image description here

I was able to get 11c pretty easily. What I thought was the intersection of the same uncountable set, say [1,2], that is [1,2]∩[1,2] would be [1,2], a uncountable set. Via help, I was able to get understand 11a. That is if you have two uncountable sets, say (−∞,0]∩[0,∞), the intersection of those two sets would be that one value, zero, meaning it is finite countable. What I am struggling with is applying that same idea to 11b. What I thought of was having two intervals that didn't end quite at the same spot say (−∞,3]∩[0,∞) but the intersection of those would be [0, 3] which itself is a uncountable set. From 11a, what endpoints would you set on the intervals so that A ∩ B would be countably infinite?

Community
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committedandroider
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3 Answers3

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How about $[0, 1] \bigcup \{2, 3, 4, 5, \dots \}$ and $[5, 6] \bigcup \{7, 8, 9, 10, \dots\}$?

  • {2,3,4,5,…} is countable though isn't it? – committedandroider Jan 20 '15 at 22:26
  • @committedandroider Yes, but when you union it with $[0,1]$ the set becomes uncountable. – J126 Jan 20 '15 at 22:30
  • but the question is the union of two uncountable sets, not one uncountable and the other countable – committedandroider Jan 20 '15 at 22:31
  • @Bonehead Let $A = [0, 2] \cup {2, 3, 4, 5, \dots}$ and $B = [5, 6] \cup {7, 8, 9, 10, \dots}$. Then $A$ and $B$ are uncountable and $A \cap B = {5, 6, 7, 8, \dots}$ is countably infinite. –  Jan 20 '15 at 22:33
  • @committedandroider Let $A=[0,1]\cup {2,3,4,\ldots}$ and $B=[5,6]\cup {7,8,9,\ldots}$. Then both $A$ and $B$ are uncountable, but the intersection $A\cap B$ is countable. – J126 Jan 20 '15 at 22:33
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    @JoeJohnson126 Funny how we both wrote almost exactly the same sentence! –  Jan 20 '15 at 22:34
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    @OohAah When I submitted my comment, I thought, for a second, that I somehow submitted it twice. – J126 Jan 20 '15 at 22:35
  • Ahh i get it thanks guys!! Tricky problem. But @0ahAah, are you actually Gina G? – committedandroider Jan 20 '15 at 22:36
  • Yes. I have now retired from music industry and I am trying to learn some math these days. –  Jan 20 '15 at 22:37
  • @OohAah , to represent {2,3,4,5} in set builder notation, would you do {x \in\ Z | x >= 2} ? – committedandroider Jan 27 '15 at 20:55
  • You should write {x \in \Z | 2 <= x <= 5}. –  Jan 27 '15 at 21:02
  • but if you were to follow that pattern and do {y \in\ Y| 7 <= y <= 10}, wouldn't you get the empty set for the intersection? – committedandroider Jan 27 '15 at 21:22
  • Oh I see. You want to write ${2, 3, 4, 5, \dots}$ as ${x \in \mathbb{Z} : x \geq 2}$. That was correct. I thought you wanted to write ${2, 3, 4, 5}$ which is finite without the dots. –  Jan 27 '15 at 21:25
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$$ A = \bigcup_{\text{even }n} [n,n+1],\qquad B = \bigcup_{\text{odd }n} [n,n+1] $$ The intersection is the set of all integers.

Michael Hardy
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let A and B be two uncountable sets . (to show AUB is uncountable ) obviously A is contained in AUB as well B is contained in AUB . since A is an uncountable set and then AUB being the superset of an uncountable set ,is itself uncountable . we are done

Salman Akhtar
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