# Convex sets in infinite dimensional Banach spaces

I am reviewing functional analysis and getting stuck in this question.

Let $X$ be an infinite dimensional Banach space. Show that there exist convex sets $K_1, K_2$ such that $K_1\cap K_2=\emptyset, K_1\cup K_2=X, cl(K_1)=cl(K_2)=X$ where $cl(E)$ is the closure of $E$.

I have no idea about where to start. Could anyone give me some hints ? Thank you very much.

Hint: There is $f : X \to \mathbb{R}$ which is linear and unbounded.

• But the complement of the kernel is not convex. – Martin Argerami Jun 11 '16 at 16:56
• @MartinArgerami Obviously. And I never said something about the kernel ;) But starting from the kernel you can develop some idea to get desired $K_1$ and $K_2$. – gerw Jun 11 '16 at 18:34
• Can you be more explicit? – user99914 Jun 11 '16 at 19:53
• What can you say about the kernel of $f$? It has some very nice properties. Similarly, which properties does the set $\{x \in X \mid f(x) = 1\}$ have? This can be used to construct $K_1$ and $K_2$. – gerw Jun 11 '16 at 20:37
• I think I got it. You have that $\{x: f(x)=1\}$ is also dense in $X$. Now, let $K_1=\{x: f(x)\le 0\}$, $K_2=\{x: f(x)>0\}$, then $K_1,K_2$ solve our problem. – Omega Jun 12 '16 at 1:00

I have some approach, but I don't know much about Banach spaces, so I don't know if it works.

We define the sequence of convex and disjoint sets $A_k$ and $B_k$, where $A_k \cup B_k$ is a $k$-dimensional subspace, as follows:

• $A_0 = \{0\}$ and $B_0 = \emptyset$
• For every $n$, let $S_n = A_n \cup B_n$, and let $x_n \in X$ be an arbitrary vector that is not in $S_n$. Then we let $$A_{n+1} = A_n \cup \{s + \lambda x_n : s \in S_n, \lambda > 0\},\\ B_{n+1} = B_n \cup \{s - \lambda x_n : s \in S_n, \lambda > 0\}.$$

The cool part of this construction is that for every $n$, we have $A_n \in \text{cl } B_{n+1}$ and $B_n \in \text{cl } A_{n+1}$!

Now, let $A = \bigcup_n A_n$ and $B = \bigcup_n B_n$. If $X$ is countably dimensional, we should have $X = A \cup B$ (at least if we choose the $x_n$ nicely). Then, we should also have $\text{cl } A = \text{cl } B = X$.

But if $X$ is uncountably dimensional, this doesn't work that well. My idea would be to make the sequence of $A_n$ "even longer" using transfinite induction, but I'm not sure if this works.