Re: Homeomorphisms of Banach spaces are open

jane wrote:
It is not known and not difficult to prove that if
$X$ is a Banach space, then the set of homeomorphisms


of $X$ is an open subset in the operator topology of
$L(X,X)$ , the set of [b]all[/b] continuous maps from
$X$ to itself.

You need to clarify a few things. I have no idea what
the operator
topology on the set of all homeomorphisms of X is.
The notation L(X,X)
makes one suspect that we're only talking about
linear maps, but
when you say "[b]all[/b] continuous maps" it sounds
like you
may mean "all continuous maps"...

You are right, i forgot to mention some things.

So, X is a normed vector space.
L(X,X) is the set of _continuous_ linear operators
from X to itself, which becomes the normed vector
space on its own right with the usual operator
topology ||A|| = sup_||x||=1 ||Ax||.

We are interested in the set of operators A from L(X,X)
such that both A and A^{-1} exist and continuous.


I suspect it is true only when $X$ is Banach, but
can't come up with a counterexample in case $X$ is
not Banach. Could you give a counterexample in this
case ?

What sort of non-Banach X are we talking about? A
general topological
space? A normed vector space?

By non-Banach, i'm interested in normed vector space,
which are not complete.

Thanks.

Thanks.
.

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