Union,intersection and algebraic sum of convex sets is convex

Question

Proof that union, intersection and algebraic sum of convex sets in a normed linear space is a convex set.

There is no problem to show for two sets. I can't figure out how to make correct induction steps.

Answer 1

An example:

$\textbf{Defintion}$ : A topological space is a non-empty set $X$ with a collection $\tau\subset P(X)$ such that $\emptyset, X\in\tau$, $\tau$ is closed under any union and if $A,B\in\tau$ then $A\cap B\in\tau$.

How would one show here that for any $n\in\mathbb{N}$, if $A_1,\dots, A_n\in\tau$ then $\cap_{i=1}^nA_i\in\tau$ ?

Let $P(n)$ be the proposition $$\big(A_1,\dots, A_n\in\tau\big)\implies \cap_{i=1}^nA_i\in\tau$$

We need to show that $P(n)$ is true for all $n\geq2$. We use induction. For $n=2$ the proposition is true by definition. Suppose that we know that $P(n)$ is true. We want to show $P(n+1)$. Remember! $P(2)$ is still true!

So if $A_1,\dots,A_n,A_{n+1}\in\tau$, then by $P(n)$ (we assume that $P(n)$ is true, that's how induction works) we have that $\cap_{i=1}^nA_i\in\tau$. But it is also true that $A_{n+1}\in\tau$. So we have that $\cap_{i=1}^nA_i\text{ and } A_{n+1} $ are elements of $\tau$. By $P(2)$, their intersection $(\cap_{i=1}^nA_i)\cap(A_{n+1})$ is an element of $\tau$. But this intersection is precisely $\cap_{i=1}^{n+1}A_i$, so we just proved $P(n+1)$.

By induction, $P(n)$ is true for all $n\geq2$. Can you see how to apply this example in your situation?