Null spaces and projections

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I'm following Kreyszig's exposition of projections in "Introduction to Functional Analysis with applications". I'm trying to follow the proof of the following theorem (9.6-1 on pp. 486-487) regarding projections and their relationship to the following partial ordering:

Definition (Partial ordering via $\leq$) Take $T_{1},T_{2}:H\to H$ to be bounded self-adjoint linear operators on a complex Hilbert space $H$. We write $T_{1}\leq T_{2}$ iff $\left\langle T_{1}x,x\right\rangle \leq\left\langle T_{2}x,x\right\rangle $ for each $x\in H$.

Theorem Suppose that $P_{1}$ and $P_{2}$ are projections on the Hilbert space $H$. Let $P_{1}\left(H\right)$ and $P_{2}\left(H\right)$ be the subspaces onto which $H$ is projected by $P_{1}$ and $P_{2}$. Furthermore, let $\mathscr{N}\left(P_{1}\right)$ and $\mathscr{N}\left(P_{2}\right)$ be the null spaces of $ $$P_{1}$ and $P_{2}$, respectively. T.f.a.e.:

1) $P_{2}P_{1}=P_{1}P_{2}=P_{1}$;

2) $\left\Vert P_{1}x\right\Vert \leq\left\Vert P_{2}x\right\Vert ,\forall x\in H$;

3) $P_{1}\leq P_{2}$;

4) $\mathscr{N}\left(P_{2}\right)\subseteq\mathscr{N}\left(P_{1}\right)$;

5) $P_{1}\left(H\right)\subseteq P_{2}\left(H\right)$.

I'm struggling to justify Kreyszig's justification for 4) implies 5). He uses the following Lemma, which I think I understand:

Lemma The orthogonal complement $Y^{\perp}$ of a closed subspace $Y$ of a Hilbert space is the null space $\mathscr{N}\left(P\right)$ of the orthogonal projection $P$ of $H$ onto $Y$.

Indeed, to prove 4) implies 5), Kreyszig says,

"This is clear since $\mathscr N \left(P_j \right)$ is the orthogonal complement of $P_j(H)$ in $H$."

I understand this to mean that we have that $\mathscr{N}\left(P_{i}\right)=\left(P_{i}\left(H\right)\right)^{\perp}$. Hence, from 4), $$\left(P_{2}\left(H\right)\right)^{\perp} \subseteq \left(P_{1}\left(H\right)\right)^{\perp}.$$

Am I right? And if I am, how does this help us to get 5)?

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Note that

  1. For any subsets $A,B$ of $H$ we have $A\subset B\implies B^\perp\subset A^\perp$
  2. $N(P)^{\perp}=P(H)$