# $\overline V= \overline {V_1}\oplus \overline {V_2}$?

Let $V$ be a topological vector space. And $V=V_1\oplus V_2$(Algebraic Sum)

Now, consider $\overline V$, the closure of $V$. Can we say that,

$$\overline V= \overline {V_1}\oplus \overline {V_2}$$

MY WORK

Clearly, $\overline {V_1}\oplus \overline{V_2}\subset \overline{V}$.

For the otherway round, letting $x\in \overline{V}$, we have a sequence $(x_n)$ in $V$ such that $x_n\to x$.

So, for all $n$ there exist $x_{n1},x_{n2}$ s.t, $x_{n1}+x_{n2}\to x$

It is not true in general that $\overline{V_1 \oplus V_2} = \overline{V_1} \oplus \overline{V_2}$. The problem is $\overline{V_1} \oplus \overline{V_2}$ is not necessarily closed as we may construct two closed subspaces such that their sum is not closed :

Let us consider the topological space $E = \ell^1 \oplus \ell^2$ and the subspaces $V_1 = \ell^1 \oplus \{0\}$ and $V_2 = \{ (u, v) \in E, \, u = v\}$.

Now clearly $V_1 \cap V_2 = \{0\}$, let's denote $V$ their algebraic sum (we have $V = \{(u,v) \in E, v \in \ell^1\}$).

$$V = V_1 \oplus V_2$$

We can easily check that $V_1$ and $V_2$ are closed. But let us prove however that $V$ is not closed (which will imply that $V = V_1 \oplus V_2 = \overline{V_1} \oplus \overline{V_2} \ne \overline{V}$) :

Chose a sequence $v$ that is in $\ell^2$ but not in $\ell^1$ (for example, set $v = \left(\frac{1}{k}\right)_{k \ge 1}$) and a sequence of sequences $v_n \in \ell^1$ that converge to $v$ for the topology in $\ell^2$ (for example, set $v_n = v . 1_{[\!|1,n|\!]}$). Then the element $(0, v) \in E$ cannot be in $V$, but is the limit of the sequence $(0,v_n) = (-v_n,0) + (v_n, v_n) \in V$.

• may you give me $V$ explicitly – David Oct 24 '15 at 11:23
• @gloom : You can check that $V = \{(u,v) \in E, v \in \ell^1 \}$. We may think of it as $V = \ell^1 \oplus \ell^1$ but where the topology on the second component is that of $\ell^2$ (that's why $V$ isn't closed : because $\ell^1$ is not closed inside $\ell^2$). – Joel Cohen Oct 24 '15 at 15:37