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Let $X$ be a normed linear space, $M$ a linear subspace of $X$. The following proposition

If $x_0 \in X$ and not in the closure of $M$, then there exists $T \in X^{\ast}$ such that $Tx_0 \neq 0$ and $T_{|M} = 0$

is mentioned as a corollary of this version Hahn-Banach theorem:

If $S$ is a bounded linear functional on $M$, then there exists $T \in X^{\ast}$ such that $Tx = Sx$ for all $x \in M$, and $||T|| = ||S||$.

Why is this? I don't see why I can't just argue this directly. If $x \not\in \overline{M}$, then I can find a complement $W$ to $\overline{M}$ such that $X = W \oplus \overline{M}$ such that $x \in W$, a bounded linear functional $S$ on $W$ for which $Sx \neq 0$, and then extend $S$ to all of $X$ by defining $S(0,m) = 0$ for $m \in M$.

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One does not always have a complement $W$ (If it does, even the Hahn Banach Theorem is trivial). See a counterexample here.

To prove this, consider $M' = M \oplus \langle x\rangle$. This is a closed subspace of $X$. Define $T(m + ax) =a$ on $M'$ and use Hahn Banach Theorem to extend this $T$ to $X$.

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