Let $B \subseteq \mathbb{R}^2$ be a simply connected closed set. I am trying to prove the following things. It appears to be very simple, and I can "prove" by picture but actually having a not so easy time making it rigorous... Any comments are appreciated.
1) How to show $B + [-1,1]^2$ simply connected?
2) Suppose $(a,b) + [-10 L, 10L]^2 \subseteq B + [-L,L]^2$. Here $L>0$. Then $$ (a,b) + [-L, L]^2 \subseteq B. $$
No, $B + [-1,1]^2$ need not still be simply connected.
For instance, let $B$ be a circle of radius 10 centered at the origin, with a very small piece missing where it would otherwise cross the positive $x$-axis. Then $B$ is simply connected, but $B + [-1,1]^2$ will be something like an annulus, and thus have a non-trivial fundamental group.