Translational invariance clarification.

Question

The question and its solution are given below (note you have to replace $n$ in the solution by $c$):

My question is why $c$ (or $n$ as in the solution) is clearly translation invariant? could anyone explain this for me please?

Answer 1

If $r\in\mathbb R$ and $E\subseteq R$ then a bijection $E\mapsto r+E:=\{r+x\mid x\in E\}$ exists that is prescribed by $x\mapsto r+x$.

Based on this it can be concluded directly that

• $E$ has infinitely many members $\iff r+E$ has infinitely many members.
• $E$ has $n$ elements $\iff r+E$ has $n$ elements (where $n$ is a positive integer).
• $E=\varnothing\iff r+E=\varnothing$.

This comes to the same as $c(E)=c(r+E)$.

Answer 2

If $A$ is an infinite set then $\{a+x: a \in A\}$ is also an infinite set fro any real number $x$. If $A$ is a finite set then $\{a+x: a \in A\}$ is also a finite set and has the same number of elements as $A$. Hence $c(A+x)=c(A)$.