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 Noche Jan 23 at 12:34

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 2
    $\begingroup$ The fact that $B\subseteq A$ tells us that $B\cap A=B$, not $A$. $\endgroup$ – lulu Jan 22 at 22:12
  • $\begingroup$ Note: the claim in the header is clearly false...did you mean to write $A\subseteq B$? $\endgroup$ – lulu Jan 22 at 22:13
  • 1
    $\begingroup$ This is false. You cannot prove it. $\endgroup$ – Ben W Jan 22 at 22:14
  • $\begingroup$ Can you clarify your question? As stated, the statement you seek to prove is clearly wrong. Perhaps you mistyped? $\endgroup$ – lulu Jan 22 at 22:18
  • $\begingroup$ Ah thanks for pointing out the mistake. Let me quickly review the question and my proof again. $\endgroup$ – fesodes Jan 22 at 22:20

Sometimes a figure is worth 1000 words.

enter image description here

By the way, why do you need "and $x \in A$" in your title?

  • $\begingroup$ What if B is the empty set? $\endgroup$ – Jack Pfaffinger Jan 22 at 22:23
  • $\begingroup$ @JackPfaffinger: Then the statement is false and any proof that doesn't exclude the empty set is invalid. $\endgroup$ – David G. Stork Jan 22 at 22:25
  • $\begingroup$ Yeah, that's what makes me wonder if he meant $x \in B$. $\endgroup$ – Jack Pfaffinger Jan 22 at 22:26
  • $\begingroup$ Me too......... $\endgroup$ – David G. Stork Jan 22 at 22:26
  • $\begingroup$ 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. $\endgroup$ – fesodes Jan 22 at 22:36

Not the answer you're looking for? Browse other questions tagged or ask your own question.