# Counterexample for “continuous image of closed and bounded is closed and bounded” (in normed spaces).

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?

• There aren't too many well-known examples of the first one, since such a map is necessarily non-linear. For the second, James's Theorem guarantees the existence of continuous linear functionals on non-reflexive spaces which map the closed unit ball to an open (and not closed) interval in $\Bbb{R}$. – Theo Bendit May 12 '19 at 15:12
• See this for the first question. – David Mitra May 12 '19 at 15:22

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.
• And by Tietze-Urysohn you can extend any such function to a continuous function on all of $\ell^2$. – Jochen May 13 '19 at 9:59