ker(T) of linear unbounded functional $T: X\to R$ is dense subset of X

Question

Let X be a normed space, $T: X\to R$ a linear unbounded functional. I have to show that

a) ker(T) is dense in X

b) $ker(T) \subset X $

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I tried to show this with contraposition:

Assume $ker(T)$ is not dense in X.

$\exists x_0\in X$ and there exists an open ball $U(x_0)$ which is disjunct to $Ker(T)$.

$T(U-x_0)$ is convex(linear image of convex set) and symmetric ($a\in T(U-x_0) \implies -a\in T(U-x_0)$.

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my final goal is to show that f has to be bounded then... but I don't find a way to continue