Topological vector space

T

If $P (x)$ is some unary predicate (a true or false statement dependent on $x \in X$ ) then for any $z \in X,$ $z + {x \in X : P (x)} = {x \in X : P (x - z)} .$ ^{[proof 1]} So for example, if $P (x)$ denotes " $| x | < 1$ " then for any $z \in X,$ $z + {x \in X : | x | < 1} = {x \in X : | x - z | < 1} .$ Similarly, if $s \neq 0$ is a scalar then $s {x \in X : P (x)} = {x \in X : P (\frac{1}{s} x)} .$ The elements $x \in X$ of these sets must range over a vector space (that is, over $X$ ) rather than not just a subset or else these equalities are no longer guaranteed; similarly, $z$ must belong to this vector space (that is, $z \in X$ ).

Proofs

^ $z + {x \in X : P (x)} = {z + x : x \in X, P (x)} = {z + x : x \in X, P ((z + x) - z)}$ and so using $y = z + x$ and the fact that $z + X = X,$ this is equal to ${y : y - z \in X, P (y - z)} = {y : y \in X, P (y - z)} = {y \in X : P (y - z)} .$ Q.E.D. $◼$

[1] $z + {x \in X : P (x)} = {z + x : x \in X, P (x)} = {z + x : x \in X, P ((z + x) - z)}$ and so using $y = z + x$ and the fact that $z + X = X,$ this is equal to ${y : y - z \in X, P (y - z)} = {y : y \in X, P (y - z)} = {y \in X : P (y - z)} .$ Q.E.D. $◼$

[proof 1]