I started with:
Assume $A\cup B = A$, then $x\in A \cup B$.
Without loss of generality, let $x \in A$.
However, at this point I am not sure what to do.
Note: In the textbook, they use proof by contrapositive, but I want a direct proof.
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$\begingroup$ You're trying to prove something about $B$, so you should probably start with $x \in B$. And as written, you're concluding something about $x$ before you define it; you should start with "Assume $A \cup B + A$ and let $x \in ...$." $\endgroup$– user296602May 12, 2016 at 23:35
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1$\begingroup$ For any $x$, if $x\in B$, then $x\in A\cup B$, then $x\in A$. $\endgroup$– peterwhyMay 12, 2016 at 23:38
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4 Answers
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Along the lines of @T.Bongers's comment:
Let $x \in B$.
Then $x \in A\cup B$ (by definition of union).
By hypothesis $A \cup B=A$.
Therefore $x \in A$.
Since $x$ was arbitrary, we have shown that for any $x \in B$, $x$ is also an element of $A$.
It follows that $B \subseteq A$ by the definition of subset.
answered May 12, 2016 at 23:40
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$\begingroup$ Couldn't we also conclude that $A \subseteq B$? $\endgroup$– jlcvMay 12, 2016 at 23:56
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$\begingroup$ It does seem that way at first, but just because every element of $x$ is in A does not mean that there could not exist an element of $A$ which is NOT in B. For example, let A be the set of plants, and B be the set of cacti -- then a eucalyptus is in A but not B. The proof given above does not contradict this, since every x considered is a cactus. $\endgroup$ May 13, 2016 at 0:00
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$\begingroup$ @ZoomBee For a more "mathy" example, consider any non-empty set $A$ and let $B=\emptyset$. Then $A\cup B=A$ and, trivially, $B\subseteq A$, but it is not at all possible that $A\subseteq B$. $\endgroup$ May 13, 2016 at 0:15
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$\begingroup$ How do you guys approach these problems? I am having trouble with where I need to start. For example, I'm currently working on this problem Prove the following: If $A \cap B = A \cap C$ and $A \cup B = A \cup C$, then $B = C$. $\endgroup$– jlcvMay 13, 2016 at 1:32
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$\{ x: x \in A \ \text{or} \ x \in B\} = \{ x: x \in A \}$, so $x \in A \cup B \Longrightarrow x \in A$.
Thus, $\{x \in A \Longrightarrow x \in A\} \ \land \ \{x \in B \Longrightarrow x \in A\} $ implies $B \subseteq A$
answered May 12, 2016 at 23:46
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If $A\cup B=A$ , then every element of the union of $A$ and $B$ is an element of $A$. Now every element of $B$ is something because of reasons, so therefore the assumption infers that every element of $B$ would be an element of $A$. $$A\cup B=A \implies B\subseteq A$$
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More directly: Because of some statement by definition, then if $A\cup B=A$ is assumed we may infer $B\subseteq A$.
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answered May 12, 2016 at 23:51
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$A\cup B=A$ iff there does not exist an element of $B$ not in $A$ iff $B$ is a subset of $A$.
