MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let H denote Hilbert space, the space of square-summable infinite sequences of real numbers-which is infinite-dimensional and separable. Let S1,S2 denote subsets of H such that a point p of H belongs to S1 or S2 according to whether the sum of the abslolute values of the co-ordinates of p is convergent or divergent respectively. Has the topology of these subsets of H been much investigated? It is clear that the positive multiples of any point p of H remain in the same set (S1 or S2) that p itself belongs to. Since the origin of H belongs to S1 it is clear that S1 is arcwise connected. But is S2 also arcwise connected? I can see no easy way to prove this. Each point of H is a limit point of S1 and also a limit point of S2. Hence neither of these sets is either an open or a closed subset of H. Is it possible that S1 and S2 are not even Borel subsets of H? I started thinking about these sets in the course of trying to come up with a series of positive rational numbers whose convergence or divergence might be undecidable in Peano's Arithmetic or even in ZFC.

share|cite|improve this question
up vote 6 down vote accepted

The unit ball of $\ell_1$ is weakly compact in $\ell_2$ and closed in $\ell_2$, so $S1$ is $F_\sigma$ and $S2$ is $G_\delta$.

ADDED 10/22/11: $S2$ is also arcwise connected. Give $x_i$ in $S2$ for $i=1,2$ you can choose a partition $A \cup B$ of the natural numbers so that $1_A x_i$ and $1_B x_i$ are in $S2$ for $i=1,2$. The straight line paths from $x_1$ to $1_A x_1 +1_Bx_2$ and from $1_A x_1 +1_Bx_2$ to $x_2$ stay in $S2$.

share|cite|improve this answer
Thanks alot for providing such clear and complete answers to all of my questions. Your proof of the arcwise connectedness of S2 is really neat. – Garabed Gulbenkian Oct 24 '11 at 22:07

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.