Wikipedia (and my teacher) state Riesz's lemma as follows:
Let $X$ be a normed linear space and $Y$ be a subspace in $X$. If there exists $0 < r < 1$ such that for every $x\in X$ with $||x|| =1$ , one has $d(x, Y) < r$, then $Y$ is dense in $X$.
Wikipedia then goes on to say
In other words, for every proper closed subspace Y, one can always find a vector x on the unit sphere of X such that d(x, Y) is less than and arbitrarily close to 1.
This is in fact the way Riesz's lemma is stated in several other places (e.g. appendix B of the book A Taste of Topology by Volker Runde).
Now, I don't see how these two statements are so quickly equivalent. The second one seems to be the contrapositive of the first one.
But this would mean that "non dense subspace" is the same as "proper closed subspace".
This doesn't seem to be true: I thought, for instance, of $C([0,1])$, the space of continuous functions on $[0,1]$ with the supremum norm, and of the subspace of differentiable functions. It's a dense subspace which is not closed.
What is (obviously) true is that proper closed subspaces are not dense.
So, correct me if I'm wrong, but it seems that to say "in other words" is innacurate, as the second statement is stronger than the first one!
Is my reasoning correct?
Best Answer
The statements are equivalent because not being dense is equivalent to the closure being proper, and the distance to a set is the same as the distance to its closure. The second statement is therefore a reformulation of the contrapositive of the first.
More explicitly, the second statement follows directly from the contrapositive of the first because, as you mentioned, proper closed subspaces are not dense. The contrapositive of the first statement follows from the second because if $Y$ is not dense, then $\overline{Y}$ is a proper closed subspace, so by the second statement there are unit vectors in $X$ whose distances to $\overline{Y}$, and therefore to $Y$, are arbitrarily close to $1$.