Show set equality.

Question

I know that to show set equality you must show that the two sets are subsets of each other. I'm having trouble showing that S\T is a subset of (S U T) given the assumption that T is a subset of S.

I let y be an arbitrary element of S\T which implies that y is an element of S and y is not an element of T. If y is not in T then that implies y is in S or it might not be in S, but I don't know how to go any further, to prove that y is in (S U T) / (S ^ T).

Any help is appreciated.

Sorry for the formatting.

Answer 1

If $T\subset S \implies T\bigcup S=S $, and $T\bigcap S=T$

Answer 2

First of all, we can see that this is true because, by definition, $S \oplus T$ is the set of elements in $S$ or $T$ but not both, and $T$ is a subset of $S$ so $S \oplus T$ is just the set of elements that are in $S$ but aren’t in $T$ (because being in $T$ implies being in $S$).

Considering this, we can write a formal proof.

Assume $x \in S \oplus T$. Then $x \in (S \cup T) \setminus (S \cap T)$, so $x \in S$ or $x \in T$, and $x \not \in S \cap T$. We know that $x \not \in T$ because otherwise we would have $x \in S$ (because $T \subseteq S$) and therefore $x \in S \cap T$. So $x \in S \setminus T$, and hence $S \oplus T \subseteq S \setminus T$.

Now assume $x \in S \setminus T$. So $x \in S$ and $x \not \in T$. Clearly, $x \in S \cup T$ and $x \not \in S \cap T$, so $x \in (S \cup T) \setminus (S \cap T) = S \oplus T$. So $S \setminus T \subseteq S \oplus T$.

It follows that $S \oplus T = S \setminus T$.

Answer 3

I let $y$ be an arbitrary element of $S\setminus T$ which implies that $y$ is an element of $S$ and $y$ is not an element of $T$.

So far so good.

If $y$ is not in $T$ then that implies $y$ is in $S$ or it might not be in $S$,

What's happening here? You've just concluded that $y$ is in fact in $S$ -- simply because $y$ is not in $T$ doesn't mean you have to forget whether it is in $S$.

It looks like you've somehow confused yourself into treating "and" as an "or" here. But when you know "$y\in S$ and $y\notin T$", this means that you know both of $y\in S$ and $y\notin T$ at the same time.

You could instead continue with something like:

• Since $y\in S$ we have $y\in(S\cup T)$. And since $y\notin T$ we have $y\notin(S\cap T)$. So when we get to $(S\cup T)\setminus(S\cap T)$, we know that $y$ is in the set we're removing something from, and it is not one of the elements we remove. So $y$ is still a member of the difference set.

Note that this direction doesn't use the assumption $T\subseteq S$. You will need that assumption for the $S\oplus T \subseteq S\setminus T$ direction, though.