-1
$\begingroup$

You think this is the right way to solve the question?

question:

Prove if $A\subseteq$B Then A$\cap$B $\subseteq$ B$\cap$C

solve:

A=B

(x$\in$ A)$\supseteq$ (x$\in$ B) $\implies${(x$\in$ A) $\wedge$ (x$\in$ C) , (x$\in$ B) $\wedge$ (x$\in$ C)} $\implies$ (x$\in$ A) $\wedge$ (x$\in$ C) $\subseteq$ (x$\in$ B) $\wedge$ (x$\in$ C) $\implies$ A$\cap$B $\subseteq$ B$\cap$C

$\endgroup$
3
  • 1
    $\begingroup$ I think what OP wants to do is prove that if $A\subseteq B$ implies $A\cap B\subseteq B\cap C$, then $A=B$. $\endgroup$ Mar 4, 2013 at 11:55
  • $\begingroup$ It is very unclear what is being asked here. flashdesign, maybe you should try to explain your question more $\endgroup$ Mar 4, 2013 at 12:08
  • $\begingroup$ After we know what you tried to ask, would you be so kind to rephrase the question in something less confusing ? $\endgroup$ Mar 4, 2013 at 13:23

2 Answers 2

2
$\begingroup$

If the question is what I think it is (as indicated in my comment on the question), then what you want to do is assume $A\subsetneq B$, let $x$ be in $B$ but not in $A$, let $C=\{{x\}}$, and take it from there.

$\endgroup$
5
  • $\begingroup$ yes.Thanks for your help. $\endgroup$
    – Software
    Mar 4, 2013 at 12:06
  • $\begingroup$ If the question was what I thought it was, then it's not what azimut thought it was --- so I don't understand why you accepted azimut's answer. $\endgroup$ Mar 4, 2013 at 12:36
  • $\begingroup$ excuse me.You are right. $\endgroup$
    – Software
    Mar 4, 2013 at 13:22
  • $\begingroup$ Now do you think my answer to this question is true? $\endgroup$
    – Software
    Mar 4, 2013 at 13:26
  • $\begingroup$ There are problems right at the start. $x\in A$ and $x\in B$ are sentences, not sets, so it makes no sense to say one of them contains the other. There's also a problem at the very end: you are concluding $A\cap B\subseteq B\cap C$, but what the problem asks you to prove is $A=B$. I haven't looked at anything in the middle. $\endgroup$ Mar 4, 2013 at 22:09
1
$\begingroup$

Hint: How do you proof a statement about a set inclusion? You start with an arbitrary element of the "smaller" set, and then you have to show that it is contained in the bigger set. So try to fill the following frame:

Let $x\in A\cap B$. Then ... So $x\in B\cap C$.

EDIT: I have typed this without thinking. The claimed statement is wrong without further preconditions. (Counterexample: $A = B = \{1\}$, $C = \emptyset$.)

$\endgroup$
4
  • $\begingroup$ who upvoted an answer which says itself, that it is wrong ? $\endgroup$ Mar 4, 2013 at 11:53
  • $\begingroup$ @Dominic, not me, but the EDIT is useful (except I think the question's misunderstood). $\endgroup$ Mar 4, 2013 at 11:57
  • $\begingroup$ @Dominic: The answer is not wrong. It gives a correct method to show the inclusion of two sets. However, the statement claimed in the question is wrong, so obviously the correct method won't work here (since you cannot correctly prove something that is untrue). $\endgroup$
    – azimut
    Mar 4, 2013 at 12:22
  • $\begingroup$ @azimut I meant the wrong not like that something in your answer is wrong, but that your answer can't be the answer to the question which has been asked $\endgroup$ Mar 4, 2013 at 13:25

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .