Set theory union and intersection problem Let us consider the following three statements


*

*$A \subset B$ 

*$A \cup  B = B$

*$A  \cap B = A$


It is intuitive that these three are equivalent. I want to prove $.1 \implies  2.$ and $1.\implies 3.$


*

*$1. \implies  2.$:
$$A  \cup  B = \{x \mid x \in A  \lor x \in B   \}$$
From 1. we can write 
$$A  \cup  B = \{x \mid x \in B \lor x \in B \} \implies A \cup B = \{x \mid x \in B \} = B$$


I followed the same method for $1. \implies 3.$ and ended up with wrong conclusion as follows


*

*$1. \implies 3.$:
$$A  \cap  B = \{x \mid x \in A \land x \in B \}$$
From 1. we can write 
$$A  \cap  B = \{x \mid x \in B  \land x \in B \} \implies A  \cap  B = \{x \mid x \in B \} = B$$


Please answer me where I went went wrong. I used $x \in A \implies x \in B$ in both the cases.
 A: Your proof of $1\implies 2$  isn't quite complete.. It only proves that if $x\in A\cup B$, then $x\in B$, in other words, you only prove that $$A\cup B\subseteq B.$$
Naturally, the other direction, i.e. $B\subseteq A\cup B$ is true, and together, this means $B=A\cup B$.

However, when you apply the same logic to the second proof, you again only prove that $A\cap B\subseteq B$, but the other direction, in this case, is not obvious (and in fact not true!)

Basically, your main problem is that you think writing $x\in A$ and $x\in B$ is equivalent, but it isn't. You hurried too much in your proof, and it blew up because not every step was well thought over.
A: So we have that $\;A\subset B\iff \;\forall\,a\in A\;,\;\text{then also}\;a\in B$ . I can't see why you think this implies that $\;A\cap B=\{x\in B\wedge x\in B\}\;$...Perhaps making a simple, down to Earth argument can help here: If all men are persons, this does not mean that all elements that are men and also persons is the same as all the persons, which is what you wrote.
What is true is that $\;A\subset B\implies\;\forall\,x\in A\,,\,\,\text{then also}\,x\in B\;\text{and also}\;x\in A\implies x\in A\cap B\;$ , and from here we get $\;A\subset A\cap B\;$ . As the other inclusion is trivial you thus get (3)
A: In the direction $1.\implies 3.$: the mistake is in 

From 1. we can write $$A\cap B=\{x\mid x\in B\land x\in B\}$$

This does not follow from 1. When using that $$A\subset B \implies \{x \in A \implies x \in B\}\tag1$$ you do not get equality anymore but inclusion: $$\{x\mid x\in A\land x\in B\}\overset{(1)}\subseteq\{x\mid x\in B\land x\in B\}=B$$ and again $$\{x\mid x\in A\lor x\in B\}\overset{(1)}\subseteq\{x\mid x\in B\lor x\in B\}=B$$ So, actually, your other direction $1. \implies 2.$ has a mistake as well. From 1. indeed follows, that $$A\cup B\subseteq B, \quad \text{ and }\quad A\cap B\subseteq B$$ but this is not what you need.
A: To show that two sets $X$, $Y$ are equal : $X=Y$ you have to show that:
$$
\forall x \in X \Rightarrow x\in Y \quad \land \quad \forall y \in Y \Rightarrow y \in X
$$
So, for $1) \Rightarrow 3)$ we have:
$$
x\in A\cap B\iff (x\in A) \land (x\in B)\Rightarrow x\in A
$$
That prove the first part. For the second part we have:
$$
(x\in A) \land (A\subset B) \Rightarrow x\in B
$$
so:
$$
x\in A \Rightarrow (x\in A) \land (x\in B) \iff x\in A\cap B
$$
Note that to prove that the three statements are equivalent now you have to prove that $2)\Rightarrow 1)$ or $3)\Rightarrow 1)$
A: Expanding things to their full definitions helps me avoid mistakes like this. Here are the proof steps I came up with, shown as a table to highlight how they're similar (two properties in the same box are implicitly "and"ed together and gray means that it's carried along unchanged from the previous step).
$$
\begin{array}{|r|c|l|}
\hline
    A \subset B &
        \text{assumption} &
    A \subset B \\
\hline
    A \subseteq B &
        \text{weaken assumption} &
    A \subseteq B \\
\hline
    x \in A \Rightarrow x \in B &
        \text{definition (implicit $\forall x.$)} &
    x \in A \Rightarrow x \in B \\
\hline
    \begin{array}{r}
        \color{gray}{x \in A \Rightarrow x \in B} \\
        x \in B \Rightarrow x \in B
    \end{array} &
        \text{add trivial assumption} &
    \begin{array}{l}
        x \in A \Rightarrow x \in A \\
        \color{gray}{x \in A \Rightarrow x \in B}
    \end{array} \\
\hline
    (x \in A \vee x \in B) \Rightarrow x \in B &
        \text{combine implications} &
    x \in A \Rightarrow (x \in A \wedge x \in B) \\
\hline
    \begin{array}{r}
        \color{gray}{(x \in A \vee x \in B) \Rightarrow x \in B} \\
        x \in B \Rightarrow x \in B
    \end{array} &
        \text{add trivial assumption} &
    \begin{array}{l}
        x \in A \Rightarrow x \in A \\
        \color{gray}{x \in A \Rightarrow (x \in A \wedge x \in B)}
    \end{array} \\
\hline
    \begin{array}{r}
        \color{gray}{(x \in A \vee X \in B) \Rightarrow x \in B} \\
        x \in B \Rightarrow (x \in A \vee x \in B)
    \end{array} &
        \text{weaken implication} &
    \begin{array}{l}
        (x \in A \wedge x \in B) \Rightarrow x \in A \\
        \color{gray}{x \in A \Rightarrow (x \in A \wedge x \in B)}
    \end{array} \\
\hline
    (x \in A \vee x \in B) \Leftrightarrow x \in B &
        \text{two-way implication} &
    (x \in A \wedge x \in B) \Leftrightarrow x \in A \\
\hline
    A \cup B = B &
        \text{definition} &
    A \cap B = A \\
\hline
\end{array}
$$
You might notice how in "combine implications", we "and" the RHS's in the intersection proof, but we "or" the LHS's in the union proof. This might feel a little backwards (but then, for example, De Morgan's laws feel a little backwards too, right?). If we know both $P \Rightarrow R$ and $Q \Rightarrow R$, then we don't need both $P$ and $Q$ to show $R$; we only need either one of them, because we only need to apply one of the original statements to show $R$.
A similar phenomenon occurs with the "weaken implication" step: we add the extra membership proposition to the RHS in an "or" for the union proof, but add it to the LHS in an "and" in the intersection proof. We can "or" extra stuff to the RHS because we only need one of those RHS propositions to be true, and we already have one we know is true from our original implication. We can "and" extra stuff to the LHS because requiring extra preconditions (stuff we need to be true) doesn't ever make us wrong when all those preconditions are met, we just don't use the extra ones because we know which one we need to pick out and use with our original implication.
These two rules can be formally proven by using the definition of implication ($P \Rightarrow Q$ means $\neg (P \wedge \neg Q)$). However, the intuition I was trying to describe comes more from the Curry-Howard correspondence and thinking about implication as functions of "proof objects" (but that's not needed for this problem, of course; I just thought I'd mention it for anyone who's interested).
