[Math] For all sets $A$, $B$, and $C$, if $A \cap C \subseteq B \cap C$ and $A \cup C \subseteq B \cup C$, then $A \subseteq B$.

elementary-set-theorysolution-verification

Note: I am aware of the duplicate, but I would like to get my specific proof checked.

I am trying to prove the following:

For all sets $A$, $B$, and $C$, if $A \cap C \subseteq B \cap C$ and $A \cup C \subseteq B \cup C$, then $A \subseteq B$.

I found this one a bit tricky, but I think I was able to get it in the end.

Proof:

Let $x \in A \cap C$.

Therefore, $x$ is an element of both $A$ and $C$. And by the hypothesis, $x$ is an element of both $B$ and $C$.

However, the problem here is that we're saying that, for all element $x$ that are in both $A$ and $B$ (we are excluding the ones that are in $A$ but not in $C$), $x$ is also an element of both $B$ and $C$. Therefore, if I am not mistaken, this is insufficient to to prove that $A \subseteq B$.

Now let $y \in A \cup C$.

Therefore, $y$ is an element of $A$ or $C$ or both.

Case 1: Let $y$ be an element of $A$ but not $C$. Then $A \subseteq B$, since $y$ would then, by the hypothesis, also be an element of $B$, since we would have that $A = A \cup C \subseteq B \cup C = B$, since $C = \emptyset$.

Case 2: Let $y$ be an element of $C$ but not $A$. Then we have that $A \subseteq B$, since we would have that $C = A \cup C \subseteq B \cup C$, since $A = \emptyset$.

Case 3: Let $y$ be an element of both $A$ and $C$. Then it must be that all elements in $A \cup C$ are in $A \cap C$, which implies that $A = C$. And since we have that $A \cap C \subseteq B$, we have that $A \cap A = A \subseteq B$.

Therefore, $A \subseteq B$.

Q.E.D.

I would greatly appreciate it if people could please take the time to review my proof.

EDIT:

Thank you all for the enlightening feedback. I understand where I went wrong and have written a new proof, taking your advice into account.

New Proof:

Let $x \in A$.

The hypothesis states (assumes) that $A \cap C \subseteq B \cap C$ and $A \cup C \subseteq B \cup C$.

These assumptions imply two possibilities: $x \in C$ or $x \not\in C$.

Case 1: Suppose that $x \in C$.

$\therefore x \in A \cap C \subseteq B \cap C$

$\therefore x \in B \cap C$

$\therefore x \in B$

$\therefore A \subseteq B$

Case 2: Suppose that $x \not\in C$.

$\therefore x \in A \cup C \subseteq B \cup C$

$\therefore x \in B \cup C$

$\therefore x \in B$ (Since $x \in B \cup C$ and $x \not\in C$.)

$\therefore A \subseteq B$

Therefore, we have that $A \subseteq B$.

Q.E.D.

Best Answer

Many problems with your proof ....

First and foremost, the very set-up is not right. You need to assume that $x \in A$, and then show that $x \in B$ ... you never do this

But some other issues as well:

Case 1: Let $y$ be an element of $A$ but not $C$. Then $A \subseteq B$, since $y$ would then, by the hypothesis, also be an element of $B$, since we would have that $A = A \cup C \subseteq B \cup C = B$, since $C = \emptyset$.

Why would $C = \emptyset$? Just because $y$ is not in $C$? That does not follow

Case 2: Let $y$ be an element of $C$ but not $A$. Then we have that $A \subseteq B$, since we would have that $C = A \cup C \subseteq B \cup C$, since $A = \emptyset$.

Same mistake. Just because $y$ is not in $A$ does not mean there is nothing in $A$ at all

Case 3: Let $y$ be an element of both $A$ and $C$. Then it must be that all elements in $A \cup C$ are in $A \cap C$, which implies that $A = C$. And since we have that $A \cap C \subseteq B$, we have that $A \cap A = A \subseteq B$.

And a similar mistake again: just because $y$ is in both $A \cap C$ and $A \cup C$ does not mean that these two sets are the same.

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