Let A, B be two sets. Prove that $A \subset B \iff A \cup B = B$

I'm thinking of using disjunctive syllogism by showing that $\neg \forall Y(Y \in A).$ However, I'm not sure how the proving steps should proceed such that it leads me to that premise.

Edit: Hey guys, thanks for the input. FYI, I need to prove this using predicate logic.

  • $\begingroup$ Just take into account that always $\,A\subset A\cup B\,$ ... $\endgroup$ – DonAntonio Sep 11 '12 at 16:23
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    $\begingroup$ Well @uohzxela, now you know...it's important as people volunteering their time and knowledge here want to know both their effort's appreciated and whether their answers are well understood by the OP $\endgroup$ – DonAntonio Sep 11 '12 at 16:30
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    $\begingroup$ What's disjunctive syllogism (am joking)? To a mathematician, the first question is not "what tool shall I use?" It is "what's happening here?" So one draws a picture. Then the picture guides one to a proof. $\endgroup$ – André Nicolas Sep 11 '12 at 16:40

Taking the "longer" road. Let us review the definitions:

  1. $A\subseteq B$ if and only if for every $x\in A$, $x\in B$.
  2. $x\in A\cup B$ if and only if $x\in A$ or $x\in B$.
  3. $A=B$ if and only if $A\subseteq B$ and $B\subseteq A$.
  4. $P\iff Q$ means that if we assume that $P$ holds, then $Q$ must hold; and vice versa.

Now assume $A\subseteq B$. This means that for every $x\in A$ we have $x\in B$. We want to show that $A\cup B=B$.

  • So we take $x\in B$, then $x\in A$ or $x\in B$, and therefore $x\in A\cup B$.
  • Now take $x\in A\cup B$, we want to show that $x\in B$. By definition either $x\in A$ or $x\in B$.

    • If $x\in B$ we are done.
    • If $x\in A$ then by the assumption that $A\subseteq B$ we have that $x\in B$.

    Either way we have that $x\in B$.

We have shown that if $A\subseteq B$ then $A\cup B\subseteq B$ and $B\subseteq A\cup B$, which is by fact number $3$ to say $A\cup B=B$.

Now we need to assume that $A\cup B=B$, and deduce that $A\subseteq B$. So we need to show that if $x\in A$ then $x\in B$.

Take $x\in A$ to be an arbitrary element. Because $x\in A$ we have that $x\in A$ or $x\in B$, and therefore $x\in A\cup B$. The assumption was, however, that $A\cup B=B$ and therefore we have that $x\in B$ as wanted.

  • $\begingroup$ Your answer is very comprehensive and well-explained. Thank you. $\endgroup$ – uohzxela Sep 11 '12 at 16:59
  • $\begingroup$ In your last sentence, assuming that A is contained in B, then you take an arbitrary element of A, why you say that "X belongs to A OR X belongs to B". Shouldn't it be an AND? Because A is assumed to be contained in B. Please explain, I'm learning $\endgroup$ – JorgeeFG Oct 13 '15 at 1:14

Let $A\subset B$. Since $B\subset B$, we have $A\cup B\subset B$. Clearly, $B\subset A\cup B$. Hence $A\cup B=B$.

Let $A\not\subset B$. Then there is some $x\in A$ with $x\not\in B$. Clearly, $x\in A\cup B$. Hence $A\cup B\neq B$.

  • $\begingroup$ Good Will Hunting, good! $\endgroup$ – Graphth Sep 11 '12 at 16:54
  • $\begingroup$ Short and sweet! Thank you! $\endgroup$ – uohzxela Sep 12 '12 at 2:12

Here are the (very straightforward) first steps you should have thought of beginning with:

In one direction, suppose $A\subseteq B$: then $A\cup B\subseteq B\cup B\dots$

In the other direction, suppose $A\cup B=B$: then $A\subseteq A\cup B\subseteq\dots$


For the first implication a draw can help us: enter image description here

and now it is obvious.

Now conversely, we have:

$A \cup B =B \Rightarrow$ $A \cup B \subset B \tag{1}$ and $B \subset A \cup B\tag{2}.$ We need only the relation $(1)$ which help us to conclude that: $A \subset B.$


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