Let $I:=[\alpha,\beta]$ a perfect interval, then $\mathcal S(I,E)\subset B(I,E)$

Question

There is a proof to show that the set of jump continuous functions (in a perfect interval $I$) is a subset of the bounded functions on that interval.

A jump continuous function is characterized by the existence of all the laterals limits of all it points.

The notation: $\mathcal S(I,E)$ is the set of $E$-valued jump continuous functions in $I$, and $B(I,E)$ is the set of $E$-valued bounded functions in $I$, where $E$ is a Banach space.

This is the given proof:

The problem here is that I dont see any contradiction... clearly $(y_n)$ is a different sequence than $(x_n)$, so it dont follow that if $(y_n)$ is bounded then $(x_n)$ is also bounded, at least I dont see why it must follow.

Some clarification will be appreciated, thank you.

Answer 1

Ok, I think I understand it. I will add this answer just for the record: the key part is when in the proof its said that:

Because every convergent sequence is bounded...

This is equivalent to say that the set of cluster points of the sequence $(x_n)$ is bounded, what imply that $(x_n)$ is bounded.