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Simple Set Theory Proof

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Hi - I'm taking a linear algebra course and the first few chapters are on set theory. I'm having trouble with a very simple proof and I'd like some help on it.

Let $$S$$ and $$T$$ be groups. Given $$T \subseteq S$$ prove:

a) $$S \cap T = T$$

b) $$S \cup T = S$$

I know these are very simple proofs but please keep in mind (and I know it's ridiculous) I was never taught proofs in high school, so the concept of proving something quite obvious is new to me.

Here's my shot at it:

a) Let $$x \in T$$

- And since $$T \subseteq S$$, then $$x \in S$$

- Which means $$x \in S \cap T$$

- And of course, $$S \cap T \subseteq S$$

- And since $$S \cap \{ x | \{x\} \subseteq S\} = \{x | \{x\} \subseteq S\}$$

- It is implied that $$S \cap T = T$$

b) Let $$x \in T \ and \ y \in S$$

- Since $$S \cap T = \{x,y \ | \ x \in T \ , \ y \in S \}$$ and it is given that $$T \subseteq S$$

- And we already know that $$x \in T \Rightarrow x \in S$$

- Then the previous statement can be rewritten as $$S \cup T = \{x,y \ | \ x \in S , y \in S \}$$

- Which obviously implies that in fact $$S \cup T = \{ x \ | \ x \in S\}$$

- So $$S \cup T = S$$

Can anyone tell me if these proofs are correct and if not, explain them in similar terms so I can improve my strategies?

Sep 29, 2015