Difference of Unions is Subset of Union of Differences

Theorem

 * $$\forall n \in \N^*: \bigcup_{i=1}^n S_i - \bigcup_{i=1}^n T_i \subseteq \bigcup_{i=1}^n \left({S_i - T_i}\right)$$

Proof
Proof by induction:

For all $$n \in \N^*$$, let $$P \left({n}\right)$$ be the proposition $$\bigcup_{i=1}^n S_i - \bigcup_{i=1}^n T_i \subseteq \bigcup_{i=1}^n \left({S_i - T_i}\right)$$.


 * $$P(1)$$ is true, as this just says $$S_1 - T_1 \subseteq S_1 - T_1$$.

Basis for the Induction

 * $$P(2)$$ is the case $$\left({S_1 \cup S_2}\right) - \left({T_1 \cup T_2}\right) \subseteq \left({S_1 - T_1}\right) \cup \left({S_2 - T_2}\right)$$.

$$ $$ $$ $$

This is our base case.

Induction Hypothesis

 * Now we need to show that, if $$P \left({k}\right)$$ is true, where $$k \ge 2$$, then it logically follows that $$P \left({k+1}\right)$$ is true.

So this is our induction hypothesis:
 * $$\bigcup_{i=1}^k S_i - \bigcup_{i=1}^k T_i \subseteq \bigcup_{i=1}^k \left({S_i - T_i}\right)$$.

Then we need to show:
 * $$\bigcup_{i=1}^{k+1} S_i - \bigcup_{i=1}^{k+1} T_i \subseteq \bigcup_{i=1}^{k+1} \left({S_i - T_i}\right)$$.

Induction Step
This is our induction step:

$$ $$ $$ $$

So $$P \left({k}\right) \implies P \left({k+1}\right)$$ and the result follows by the Principle of Mathematical Induction.

Therefore $$\forall n \in \N^*: \bigcup_{i=1}^n S_i - \bigcup_{i=1}^n T_i \subseteq \bigcup_{i=1}^n \left({S_i - T_i}\right)$$.