Counterexample for "continuous image of closed and bounded is closed and bounded" (in normed spaces).

Question

It's well known that:

1. If $X$ is a finite-dimensional normed space, $C$ is a closed and bounded subset of $X$ and $f:C\subset X\to X$ is continuous, then $f(C)$ is closed and bounded.

2. If $X$ is any normed space, $C$ is a compact subset of $X$ and $f:C\subset X\to X$ is continuous, then $f(C)$ is compact.

In the finite-dimensional case, "compact" is the same as "closed and bounded". Therefore, Item 1 is a particular case of Item 2.

Question:

Item 2 does not hold with "compact" replaced by "closed and bounded", right? What are the standard counterexamples? More precisely:

• What is an example of a Banach space $X$, a closed and bounded subset $C$ of $X$ and a continuous function $f:C\subset X\to X$ such that $f(C)$ is not bounded?

• What is an example of a Banach space $X$, a closed and bounded subset $C$ of $X$ and a continuous function $f:C\subset X\to X$ such that $f(C)$ is not closed?

Answer 1

Let $X = \ell^2$, and $C$ an orthonormal basis. Then $C$ is closed and bounded, but it is a discrete set, so any function on it is continuous. In particular, you can map it continuously to any sequence, which may be unbounded or non-closed.