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Let $A = \{\{0\}\}$ and $B = \{0\}$. Which of the following statements are true and which are false? Justify each of your answers.

  1. $|A| = |B|$ (10 marks)
  2. $A \cap B = \emptyset $ (10 marks)
  3. $A \cap \mathcal{P}(B) = \emptyset $ (10 marks)

*$\mathcal{P}(B)$ denotes te power set of $B$.

L F
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scaphrax
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  • To say the answer, 1, 2 is true and 3 is false. – Hanul Jeon Oct 24 '15 at 13:04
  • $10$ marks? Are these points? – user190080 Oct 24 '15 at 13:07
  • I don't know how to solve it, why is it true for 1 and 2 and 3 is false? Can you please explain? The marks are points yes – scaphrax Oct 24 '15 at 13:09
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    Your verbatim recitation of assigned problems suggests that not only do you not "know how to solve it", you don't know how to read it. If you are having a difficulty understanding the notation of sets, it would be better to phrase a question about that (using your own words). – hardmath Oct 24 '15 at 13:13
  • @scaphrax Please follow up hardmath's advice if you want to prevent your question from being closed! :) – Yes Oct 24 '15 at 13:40
  • Why are their TWO upvotes? –  Oct 24 '15 at 15:50

3 Answers3

2

If $A = \{ \{ 0 \} \}$, then $A$ has exactly one element; if $B = \{ 0 \}$, then $B$ has exactly one element; so $|A| = |B|$. Moreover, we have $0 \notin A$ and $\{ 0 \} \notin B$, so $A \cap B = \varnothing$. Finally, note that $P(B) = \{ \varnothing, B \}$ and $A = \{ B \}$; then $P(B) \cap A = B \neq \varnothing$.

Yes
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2
  1. $|A| = |B|$

What is the cardinal of $A$? $A$ only has one element, $\{0\}$, so $|A|=1$. $B$ has one element, $0$, so $|B|=1$

We have $|A|=|B|$

  1. $A ∩ B = ∅$

Which element are at the same time in $A$ and $B$? $\{0\}$ is not in $B$, and $0$ is not in $A$, so $A ∩ B = ∅ $is true

3.$ A ∩ P(B) = ∅$

What is $P(B$)? $p(B)= \{∅,\{0\}\}$ we have $A ∩ P(B) = A \neq ∅$

Baconaro
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1

Rewrite $A$ as $\{B\}$. It is clear that 1 is true: $\lvert A\rvert =1=\lvert B\rvert$.

2 is true since $A$ and $B$ have each one element, and these elements are distinct.

3 is false, since $\mathcal P(B)=\{\varnothing, B\}$ and $B\in A$. Actually, $A\subset \mathcal P(B)$.

Bernard
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