A variant of Baire Category Theorem states that if a complete metric space is a countable union of closed sets then at least one of the sets has a non-empty interior.

In my study I have a Banach space which is a countable union of closed sets $F_n$, with the property that for every $x \in F_n$, if $\lambda \in \Bbb{K}$ with $| \lambda| \geq 1$ then $\lambda x \in F_n$. I would need to find a set out of these which contains a circle(not necessarily a disk) around the origin(not necessarily centered in the origin), but containing the origin in its interior. I know this is almost impossible using only these conditions, but maybe I can find more properties of the sets $F_n$ before leaving this lead.

My question is like this:

Are there any theorems similar to Baire Category theorem which can prove something like this? Even if there is no theorem that directly solves the problem above, I am interested in any exotic variant of Baire Category theorem, in which maybe the sets $F_n$ have additional properties.

References are welcome. Thank you.


Not even true for $\mathbb{R}^n$ for $n>1$. Let $v$ be a non-zero vector with rational co-ordinates, and let $F_v = \{w: |w \cdot v| \geq |v|^2\}$. Let $F_0 = \{0\}$.

Then $\{F_v\}$ covers $\mathbb{R}^n$, but there is no sphere containing $0$ in any $F_v$, since no vectors perpendicular to $v$ are in $F_v$.

Additional comments: I think it's wrong to consider your goal as being somehow related to the Baire Category Theorem. BCT is inherently purely topological, while what you are looking for is something geometrical, including your desire for a circle.

If you have circle in $F_n$ that contains $0$ in the interior, you can find a circle in $F_n$ which is centered on $0$.

Obviously, for $F$ to contain a circle around $0$, you need, at minimum, for each $v$, a $\lambda > 0$ such that $\lambda v \in F$. That's not sufficient, however.

  • $\begingroup$ I know it isn't true like this. I asked if there exists another variant of Baire Theorem, which might have a result similar to this. $\endgroup$ – Beni Bogosel Oct 20 '11 at 21:17

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