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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.

Chill2Macht
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jlcv
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  • 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 $x \in ...$." –  May 12 '16 at 23:35
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    For any $x$, if $x\in B$, then $x\in A\cup B$, then $x\in A$. – peterwhy May 12 '16 at 23:38

4 Answers4

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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.

Chill2Macht
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  • Couldn't we also conclude that $A \subseteq B$? – jlcv May 12 '16 at 23:56
  • 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. – Chill2Macht May 13 '16 at 00:00
  • @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 $A\subseteq B$. – Daniel W. Farlow May 13 '16 at 00:15
  • 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 $B = C$. – jlcv May 13 '16 at 01:32
  • http://math.stackexchange.com/a/555663/327486 – Chill2Macht May 13 '16 at 03:12
  • @ZoomBee The statement is not symmetric or "indifferent" between A and B. Similarly the proof is not symmetric or indifferent. in this manner. – Jacob Wakem Jun 10 '16 at 15:39
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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 $B \subseteq A$

MathematicsStudent1122
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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 $A$.

ervx
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Jacob Wakem
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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 $A$. $$A\cup B=A \implies B\subseteq A$$

$\Box$

More directly: Because of some statement by definition, then if $A\cup B=A$ is assumed we may infer $B\subseteq A$.

$\Box$

Graham Kemp
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