How can I show that $T(O)$ is not open in $Y$? where we have that $O = [{x \in X : ||x||_X < 1}]$.
$X:[ {x = (x_1, x_2, x_3, . . .) : x_n ∈ \mathbb{K}, \sum_{n=1}^{\infty}|x_n|< \infty}]$ and $||x||_X = \sum_{n=1}^{\infty}|x_n|$
$Y:[ {x = (x_1, x_2, x_3, . . .) : x_n ∈ \mathbb{K}, \sum_{n=1}^{\infty}|x_n|< \infty}]$ and $||x||_Y = sup_{n \in \mathbb{N}} |x_n|$
I was personally thinking of using the Open Mapping theorem. If $X$ and $Y$ are Banach spaces and $T : X \rightarrow Y$ is a surjective continuous linear operator, then T is an open map.
I have easily proven that $T$ is surjective and bounded.
I was thinking of proving that $0$ isn't an interior point by assuming there is a $Y$ ball around $0$ contained in $O$ and trying to find a contradiction by finding an element in the ball that's not in $T(O)$,by using a sequence of the form m(0,0,...,1,0,0,...) but this leads me nowhere... Also why does this not contradict the Open Mapping theorem?
Any help would be grateful. Thanks in advance.
It's clear that $O$ is open in $X$. So if the open mapping theorem applied, then $T(O)$ would have to be open in $Y$. Since the exercise asks you to prove that it's not, clearly the open mapping theorem doesn't apply. Since $T$ is trivially surjective, one of the spaces involved is not Banach. (One of them is carrying the wrong norm, but you should show this in detail.)
To show that $T(O)$ is not open in the space $Y$, you need to show that for no $\epsilon > 0$ we have $B(0,\epsilon) \subset T(O)$ (i.e. that 0 is not an interior point). Note that $T(O)$ is just $O$, but the points have a different norm. Can you find an element $y$ for which the $Y$-norm of $y$ is less than $\epsilon$ (so $y \in B(0,\epsilon)$), but for which the $X$-norm is greater than 1 (so $y \notin T(O)$)?
The element $m \cdot e_k$ you're suggesting would have the same norm in both spaces. You need to look for an element where the two norms disagree.