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Simple proofs for these are pretty straight forward such as proving if two sets are equal then they are subsets of each other or if you want to show one set is a subset of the other just show that there exists an element in one set and show that they are in the other but when you add in cross and implications it gets kind of tricky.

For example,prove or disprove if the statement is true or false. if $A\cap C \subseteq B \cap D \implies A \subseteq B$

My proof, Given $x\in A\cap C$ $\implies x\in A$ and $x \in C$ if $x \in A$ then $x \in b \cap D$ since $A\cap C \subseteq B \cap D $

then $x\in B\cap D $ then $x \in B$ and D since x is in A and B then they could be subsets but it also means that A=B therefore this statement should be false?

Can someone critique my proof or tell me if this isn't the way to go about this proof? Thanks in advance!

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  • $\begingroup$ What if $x \in A$ but $x$ is not in $C$? You're not given any information on what happens in this case. Try to think of some examples. $\endgroup$
    – Dorebell
    Commented Mar 17, 2015 at 4:11
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    $\begingroup$ but if $x\in A \cap C$ then x also has to be in C $\endgroup$
    – Jared
    Commented Mar 17, 2015 at 4:18

2 Answers 2

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Something more concrete.

Let $A=[0,1]$,

$C=[0.5,1.5]$,

$B=D=[0.4,1]$.

Then $A\cap C=[0.5,1]$ and $B\cap D=[0.4,1]$. So $A\cap C \subset B\cap D$. But $A$ is certainly not a subset of $B$.

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  • $\begingroup$ I came across a similar example, but i was wondering how were you able to see it so quickly? $\endgroup$
    – Jared
    Commented Mar 17, 2015 at 4:19
  • $\begingroup$ because logically it seems to make sense unless there is a mistake in my proof $\endgroup$
    – Jared
    Commented Mar 17, 2015 at 4:20
  • $\begingroup$ Well this is a counter example. So there is no 'proof' going on. You should be able to see there is something fishy going on given that $A\cap C=C\cap A$, so then should that also mean that $C\subseteq B$? It's then just trying to construct a counterexample. $\endgroup$
    – beedge89
    Commented Mar 17, 2015 at 5:07
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Let $A,B$ be arbitrary, let $C$ and $D$ be the emptyset. We have $A\cap C=B\cap D=\emptyset$. But we cannot conclude $A\subseteq B$. Hence it is false.

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