Prove that $\; U\setminus A = U \iff A=\emptyset\; $ where $U$ is the universe (False proof of the opposite).

Question

I think I might have made a false proof for the following problem but I can´t seem to find where the flaw is.

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Prove that $\displaystyle \; U\setminus A = U \iff A=\emptyset\; $ where $U$ is the universe.

My atempt

$$U\setminus A=U\iff A=\emptyset\tag{1}$$ $$\iff [(x\notin A\iff x\in U)\iff A=\emptyset]\tag{2}$$ $$\iff [(x\notin A\iff True)\iff A=\emptyset]\tag{3}$$ (this is clearly true since the $\emptyset$ is a set such that $x\notin\emptyset\iff True$ (Thus $A=\emptyset$). But, what I want to achive is arrive at the conlsuion that (1) is equivalent to $True$ using nothing but algebraic manipulation) $$\iff [(x\in A\lor True)\land (x\notin A\lor False)\iff A=\emptyset]\tag{4}$$ $$\iff [x\notin A\iff A=\emptyset]\tag{5}$$ $$\iff[x\notin A\iff (x\in A\iff x\in\emptyset)]\tag{6}$$ $$\iff [x\notin A\iff (x\in A\iff False]\tag{7}$$ $$\iff [x\in A\iff (x\notin A\lor False)\land(x\in A\lor True)]\tag{8}$$ $$\iff [x\in A\iff x\notin A\land True]\tag{9}$$ $$\iff [x\in A \iff x\notin A]\tag{10}$$ $$\iff False\tag{11}$$ $$\therefore (U\setminus A=U\iff A=\emptyset)\iff False\tag{12}$$

I proved the oposite of what I'm meant to prove. Where did I go wrong?

(please correct tags if wrong).

Answer 1

It should be simple.

If $a \in A$ then $a \not \in U\setminus A$. But everything is in $U$.

So: If $A$ is not empty. Then there exists an $a\in A\subset U$. So $a \not \in U\setminus A$ while $a \in U$. So $U \ne U\setminus A$.

And if $A=\emptyset$ then $U\setminus A=U\setminus \emptyset =U$.

....

Or if you don't want to do contrapositive proof or want more detail as to why $K \setminus \emptyset = K$ always:

Suppose $U\setminus A = U$. Then for any $a \in A$ then $a\not \in U\setminus A$ and $a\not \in U$. But everything is in $U$ so there can not be any $a\in A$. So $A$ is empty.

Suppose $A=\emptyset$. Then if $U\setminus A = \{x\in U|x \not \in \emptyset\}=\{x\in U\}\cap \{x\not \in \emptyset\}=U\cap \emptyset^c$. But as $\forall x; x\not \in \emptyset$ we have $\emptyset^c = U$. so if $A=\emptyset$ then $U\setminus A = U\setminus \emptyset = U \cap \emptyset^c = U\cap U = U$.

Answer 2

First, note that the assertion that "$U$ is the universe" implies that $A\subseteq U$. Second, using the definition of set complement, note that $A=\emptyset$ implies that $U\setminus A=U$, and $U\setminus A\subseteq U$. So, all you need to prove is that $A\neq\emptyset$ implies that $U\setminus A \not\supseteq U$. But if $A\neq\emptyset$, there must be some $B\in A\subseteq U$, and so $B\not\in U\setminus A$.