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Let $A$ and $B$ be sets. Let $A\cup B\subseteq A\cap B.$ Prove that $A\subseteq B.$

My understanding of this question is that all the elements of set $A$ that intersects with set $B$ exists in the union of sets $A$ and $B$ and because in order for the elements to intersect, the elements must exist in both sets $A$ and $B$. Therefore set $A$ is a subset of $B$.

My question is how do I write the proof down so that it is an acceptable answer? Also does this mean set $B$ is a subset of $A$?

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$$A\subseteq A\cup B\subseteq A\cap B \subseteq B.$$
Similarly $B\subseteq A.$ Hence $A=B$

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    $\begingroup$ Where does that $=$ come from in the first line? Don't you mean $A\subseteq A\cup B \subseteq A\cap B \subseteq B$ ? $\endgroup$ – Ewan Mellor Dec 17 '15 at 22:45
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    $\begingroup$ @EwanMellor Your version would slightly more clearly be using the given assumption, but since $A\cap B \subseteq A \cup B$ is straightforward, using the equals sign is fine too. $\endgroup$ – Stan Liou Dec 17 '15 at 23:29
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Suppose, we have an element $x$ with $x\in B$, but $x$ is not element of $A$

Then, $x$ is in the union of $A$ and $B$, but not in the intersection. This is a contradiction.

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How do u proceed when you need to show one set is contained in another? Take an element in first one and somehow try to show it in second one. Here let x belongs to A, then x belongs to A union B, now using given condition we will get x belongs to A intersection B, thus x belongs to B, which is precisely you need to show.

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  • $\begingroup$ Similarly you can show B is a subset of A. $\endgroup$ – user268307 Dec 17 '15 at 19:19
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$$A \cap B \subseteq A \subseteq A \cup B $$ $$A \cap B \subseteq B\subseteq A \cup B$$

If $A \cup B\subseteq A \cap B $, then the ends are squeezed and $A = B$.

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