$c_{00}$ is closed?

Question

Let $E:=c_{00}$. Recall that$$c_{00}:=\{(x_n)_{n\in \mathbb{N}}|\exists M\geq 0\,:\,(x_n)_{n\in \mathbb{N}}=(x_1,x_2,\ldots ,x_M,0,0,0,0,\ldots )\}.$$We are equipping this with the topology given by norm$$\|(x_n)\|_\infty :=\sup \{|x_n|:n\in \mathbb{N}\}.$$ Is $c_{00}$ closed?

Answer 1

The statement is wrong.

$(A+V)\cap (B+V)=0$ for every neighbourhood $V$ of zero such that $\|(x_n)\|<\frac 1 4$ for all $x \in V$. In particular for $V=B(0,\frac 1 4)$. This is because $(a_n) \in A, (b_n) \in B, (x_n) \in V, (y_n) \in V$ imply $a_1=\frac 1 2, b_1=1$, $|x_1|<\frac 1 4$ and $|y_1|<\frac 1 4$. So we cannot have $a_1+x_1=b_1+y_1$ (since $|x_1-y_1| <\frac 1 2$ but $|a_1-b_1|=\frac 1 2$.

Answer 2

The proof of geetha is correct. I'm adding another one for maybe better understanding.

Let $a\in A$ and $b\in B$. Then the distance between them is $\|a-b\|_\infty \geq |a_1-b_1| = |\frac 12-1|=\frac 12$. This shows that any point from $A$ is far away from any point from $B$. If we take a radius $r$ that is not greater than half of that distance, i.e. $r\leq \frac 14$ then the sets $A+B(0,r)$ and $B+B(0,r)$ are disjoint. Indeed, if $c\in A+B(0,r)\cap B+B(0,r)$ then $\|c-a\|,\|c-b\|<\frac 14$ for some $a\in A,\ b\in B$. Therefore from the triangle inequality we get $\frac 12\leq \|a-b\|\leq \|a-c\|+\|c-b\|<2r\leq \frac 12$, a contradiction.