Operator Theory Textbook Question I read the following excerpt in my course textbook:

Now, I'm led to believe that $P:X\rightarrow X$ as above is bounded iff $M$ and $N$ are both closed. I understand the only if direction, but I can't put my finger on how $M$ and $N$ closed $\implies$ $P$ is bounded. Could somebody kindly provide a sketch proof? Thanks 
 A: Note that if $P$ is a projection, then $M$ and $N$ are complementary in $X$. In fact, $x = (I - P)x + Px$ for all $x\in X$, and we have $\mbox{Ker}\,P=(I-P)(X)$ and $\mbox{Im}\,P = P(X)$.
Your claim is true in Banach spaces, not in normed spaces in general. Let's see why.
Assume $X$ is Banach. Suppose $\{ x_n \} \subseteq X$ converges to $x \in X$ and that $\{ P x_n \}$ converges to some $y$. Then $x_n - Px_n = (I-P)x_n$ converges to $x - y$. It follows that $y \in M$ and $x-y \in N$, so $y = Py = Px$. Hence the conclusion follows from the closed graph theorem.
Now, suppose $X$ is a normed space, but not complete. For instance, take $X$ the subspace of $\ell^\infty$ of finetely nonzero sequences and take $\{ s_n \}$ an unbounded sequence of scalars. Define $P$ as 
$$P(a_n) = ( a_1 + s_1 a_2, 0, a_3 + s_2 a_4, 0, a_5 + s_3 a_6, \dots ) \qquad (\{a_n\} \in X).$$
Then $P$ is a projection (is linear and $P^2 = P$). Also
$$\mbox{Im}\,P = \{ \{a_n\} : \{a_n\} \in X, a_{2n} = 0 \mbox{ for each } n \in \mathbb N \},$$
and
$$\mbox{Ker}\,P = \{ \{a_n\} : \{a_n\} \in X, a_{2n-1} + s_n a_{2n} = 0 \mbox{ for each } n \in \mathbb N \}.$$
Then $\mbox{Ker}\,P$ and $\mbox{Im}\,P$ are complementary in $X$ and both are closed, but $P$ is not bounded.
A: Just from that statement (and not really having a grasp of the subject material) I would not imply this:
"If the range $M$ and the null space $N$ of $P$ are closed subspaces of (normed space) $X$, then $P$ is a bounded projection operator on $X$."
That's the converse of what's in the last sentence, as far as I can see.  The converse does not necessarily follow from the statement.  It can -- and that's the if and only if part -- but it doesn't have to.
