Theorem osumi

Description: If two closed subspaces of a Hilbert space are orthogonal, their subspace sum equals their subspace join. Lemma 3 of Kalmbach p. 67. Note that the (countable) Axiom of Choice is used for this proof via pjhth , although "the hard part" of this proof, chscl , requires no choice. (Contributed by NM, 28-Oct-1999) (Revised by Mario Carneiro, 19-May-2014) (New usage is discouraged.)

	Ref	Expression
Hypotheses	osum.1	$$⊢A\in {\mathbf{C}}_{\mathscr{H}}$$
	osum.2	$$⊢B\in {\mathbf{C}}_{\mathscr{H}}$$
Assertion	osumi	$$⊢A\subseteq \perp \left(B\right)\to A{+}_{\mathscr{H}}B=A{\vee }_{\mathscr{H}}B$$

Proof

Step	Hyp	Ref	Expression
1		osum.1	$$⊢A\in {\mathbf{C}}_{\mathscr{H}}$$
2		osum.2	$$⊢B\in {\mathbf{C}}_{\mathscr{H}}$$
3	1	a1i	$$⊢A\subseteq \perp \left(B\right)\to A\in {\mathbf{C}}_{\mathscr{H}}$$
4	2	a1i	$$⊢A\subseteq \perp \left(B\right)\to B\in {\mathbf{C}}_{\mathscr{H}}$$
5	1 2	chsscon2i	$$⊢A\subseteq \perp \left(B\right)\leftrightarrow B\subseteq \perp \left(A\right)$$
6	5	biimpi	$$⊢A\subseteq \perp \left(B\right)\to B\subseteq \perp \left(A\right)$$
7	3 4 6	chscl	$$⊢A\subseteq \perp \left(B\right)\to A{+}_{\mathscr{H}}B\in {\mathbf{C}}_{\mathscr{H}}$$
8	1	chshii	$$⊢A\in {\mathbf{S}}_{\mathscr{H}}$$
9	2	chshii	$$⊢B\in {\mathbf{S}}_{\mathscr{H}}$$
10	8 9	shjshseli	$$⊢A{+}_{\mathscr{H}}B\in {\mathbf{C}}_{\mathscr{H}}\leftrightarrow A{+}_{\mathscr{H}}B=A{\vee }_{\mathscr{H}}B$$
11	7 10	sylib	$$⊢A\subseteq \perp \left(B\right)\to A{+}_{\mathscr{H}}B=A{\vee }_{\mathscr{H}}B$$