Can a Dense Subspace Be of the 1st Category?

[redrzewski]

There's a problem from Rudin's Functional Analysis where I need to show something is a dense subspace of 1st category.

But I thought that it was the definition of dense that its closure is the whole space. Hence the closure doesn't have empty interior. So the dense subspace can't be 1st category.

Can someone clarify?
thanks

 

[jostpuur]

A set being 1st category does not mean that the interior of its closure is empty. Instead it means that the set can be written as a countable union of such sets, whose interiors of closures are empty.

For example \mathbb{Q} is dense and 1st category in \mathbb{R}.

 

[redrzewski]

Thank you for that excellent clarification.