Given the operator: $T: \ell^1 \rightarrow \ell^2$ with $Tx = x$ which of the following subsets of $\ell^1 \times \ell^2$ are closed?
$U = \ell^1 \times \{0\}$
$V = \Gamma_T$ the Graph of T
$U + V$
Defining the operator $A: \ell^1 \times \ell^2$ and $Ax = 0$ we have $U = \Gamma_A$.
$A$ and $T$ are both continuous and linear. by the closed graph theorem we have that $U$ and $V$ are both closed. is this correct so far?
I don't know how to prove or disprove that $U + V$ is closed. Thanks for any help.
Consider a sequence $$x_n =\left( 1,\frac{1}{2} , \frac{1}{3} ,...,\frac{1}{n} , 0,0,...\right)$$
Tnen $(-x_n ,0)\in U$ and $(x_n , x_n ) \in V$ but $$(-x_n ,0) + (x_n , x_n )\to (0, x)$$
where $x=\left(\frac{1}{k}\right)_{k\in\mathbb{N}}$ and the element $(0,x)\notin U+V$ so the set $U+V$ is not closed.