discontinuous linear functions

Let $H>\mathbb{Nat}$ be unlimited.

Then the linear map $T(x:{\mathbb{R}}^{\ast }):{\mathbb{R}}^{\ast }:=Hx$ is discontinuous. Why, its discontinuity is equivalent to it being unbounded. This holds in general, but this example is the germ of generality.

See my previous post for definitions of bounded and continuous.

The map $T$ is unbounded since it maps $1$ and $2$ to $H$ and $2H$, which are $H$ apart, an unlimited distance. It is also not continuous since $\frac{1}{H}$ is infinitely close to $\frac{2}{H}$, but $T(\frac{1}{H})=1$ is not infinitely close to $T(\frac{2}{H})=2$.

However, if $H$ was limited, then $T$ would be continuous. The scaling of $H$ would be bounded, and infinitely close points would stay infinitely close, AKA $T$ is continuous.

For experts and the pathologically curious

Existence of discontinuous linear maps on complete spaces is equivalent to forms of the Axiom of Choice. The wiki covers it more. From our radically elementary perspective, the issue is that H is an external object. One way of defining it as a sequence under the hood is H = [1, 2, 3, ..]. Its unassignablility (as Leibniz called it) is where the issue lies.