# Prove that — the range $R(T)$ of a bounded linear operator $T:X\to Y$ need not be closed in $Y$

Prove that the range $R(T)$ of a bounded linear operator $T:X\to Y$ need not be closed in $Y$.

• Hi, welcome to Math.SE! Please add your thoughts on how to approach the problem and we will be glad to give some hints. – gt6989b Mar 17 '14 at 20:09
• Take a compact operator from an infinite dimensional space with infinite dimensional range. – David Mitra Mar 17 '14 at 22:37

Consider the identity map $$\mbox{id} :\ell^1 \longrightarrow \ell^2 .$$
• but $l^1$ is complete , so must be closed in $l^2$ as a subspace , then how is the range space not closed ? – user228169 Feb 19 '16 at 10:56