+12 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?
