# Given $B \subseteq A$ and $x \in A$, x could also be in B [closed]

This statement should be true right? If $$B \subseteq A$$ and $$x \in A$$, $$\exists x(x \in B)$$. I was trying to prove this statement but my previous approach was wrong.

My previous approach involved misstating the property if $$B \subseteq A$$ then $$B \land A = A$$. This obviously is wrong as it should be $$B \land A = B$$.

So this is my new proof.

Given $$B \subseteq A$$, $$B \lor A = A$$. This means:

$$\forall x(x \in (B \lor A) \leftrightarrow x \in A)$$

The statement of interest is:

$$\forall x(x \in A \rightarrow x \in (B \lor A)$$

Thus if $$x \in A$$, x could be in B. Or that $$\exists x(x \in B)$$.

## closed as unclear what you're asking by lulu, max_zorn, Riccardo.Alestra, José Carlos Santos, Joel Reyes NocheJan 23 at 12:34

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• The fact that $B\subseteq A$ tells us that $B\cap A=B$, not $A$. – lulu Jan 22 at 22:12
• Note: the claim in the header is clearly false...did you mean to write $A\subseteq B$? – lulu Jan 22 at 22:13
• This is false. You cannot prove it. – Ben W Jan 22 at 22:14
• Can you clarify your question? As stated, the statement you seek to prove is clearly wrong. Perhaps you mistyped? – lulu Jan 22 at 22:18
• Ah thanks for pointing out the mistake. Let me quickly review the question and my proof again. – fesodes Jan 22 at 22:20

By the way, why do you need "and $$x \in A$$" in your title?
• Yeah, that's what makes me wonder if he meant $x \in B$. – Jack Pfaffinger Jan 22 at 22:26
• I realized that the property I used was wrong. $B \subseteq A$ means $B \land A = B$. With that said, the question should still be valid right? As by the picture provided by David, there should at least be an x that is in A and B. – fesodes Jan 22 at 22:36