# When is the kernel of a linear operator closed?

1. Jan 26, 2013

### AxiomOfChoice

If you consider a bounded linear operator between two Hausdorff topological vector spaces, isn't the kernel *always* closed? I mean, if you assume singleton sets are closed, then the set $\{0\}$ in the image is closed, so that means $T^{-1}(\{0\})$ is closed, right (since T is assumed continuous)? I keep finding these weird provisos all over the Internet (see, for instance, http://en.wikipedia.org/wiki/Kernel_(linear_algebra [Broken])) that require the target space to be finite-dimensional, and I don't see why this is a necessary hypothesis.

Last edited by a moderator: May 6, 2017
2. Jan 27, 2013

### micromass

You are right, if $T:V\rightarrow W$ is continuous and if W is Hausdorff, then $T^{-1}({0})$ is always closed.

Why the weird proviso's like finite-dimensional? Well, because the result in the wiki is an iff statement. So the statement in the theorem also says that if $T^{-1}(\{0\})$ is closed, then T is continuous. This is not true in general but only if W is finite dimensional!