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I am currently stuck attempting this question. The only way I know how to solving this question is simply via $(A \cap B)\subseteq A$ and $A\subseteq(A \cup B)$. Therefore $(A \cap B)\subseteq(A \cup B)$.

However, as mentioned in the question I need to use an element argument to prove it. Anyone able to advice/teach me how to do it?

Thanks.

Khashayar Baghizadeh
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mathtroubles
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  • Take any $x \in A\cap B$. Prove that this $x \in A \cup B$. – thanasissdr Jul 12 '15 at 16:28
  • is thanasissdr's comment in the right direction, or is there something we are missing? and what is P(U)? – JMP Jul 12 '15 at 16:48
  • @thanasissdr; does it change the question then? or would $U$ suffice? – JMP Jul 12 '15 at 16:56
  • @JonMarkPerry Basically, we have the equivalence: $$A \subseteq U \iff A \in \mathcal{P}(U).$$ – thanasissdr Jul 14 '15 at 03:39
  • @thanasissdr; okay, i see, does this mean $\mathcal{P}(\mathcal{P}(U))=\mathcal{P}(U)$? – JMP Jul 14 '15 at 05:08
  • @JonMarkPerry No, that's not the case. Let $U = {a,b,c}$. The number of elements in $\mathcal{P}(U)$ is: $\left|\mathcal{P}(U)\right| = 2^3 = 8 $. However, $\left| \mathcal{P}(\mathcal{P}(U))\right|= 2^8$. Consequently, $\mathcal{P}(U) \neq \mathcal{P}(\mathcal{P}(U)).$ – thanasissdr Jul 14 '15 at 05:25
  • @thanasissdr; okay, so if the question was 'for all A,B in P(P(U))' it would be false? – JMP Jul 14 '15 at 05:32
  • @JonMarkPerry No, it would be still valid. We can rename the set $\mathcal P( \mathcal P(U))$ to anything. E.g. $\mathcal P( \mathcal P(U)) = Z$. We still have a set. Elements in $Z$ can be anything, like elements in $U$. All of $a,b,c \in U$ could be sets. Maybe this is a little helpful. – thanasissdr Jul 14 '15 at 05:49

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$$\begin{align} x\in A\cap B &\implies x\in A \land x\in B \\ &\implies x\in A \\ &\implies x\in A \lor x\in B \\ &\implies x\in A\cup B \end{align}$$

JMP
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Rory Daulton
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  • Hi thanks for the great help. However just to clarify the reason how do we derive from x∈A∧x∈B⟹x∈A? Is it showing that (A ∩ B)⊆A. Therefore x∈A∧x∈B⟹x∈A ? Right? – mathtroubles Jul 13 '15 at 01:44
  • That is a purely logical statement, often assumed or proven in logical systems. Example: If I say "My name is Rory and I am married" you can then infer that "My name is Rory." The next line in my proof is another logical statement: You can also then infer that "My name is Rory or my name is George." I do not know which is your logical system, so I don't know what is an axiom for you and what you have derived. – Rory Daulton Jul 13 '15 at 08:56