Metamath Proof Explorer


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 ∈ Cℋ
osum.2 ⊢ 𝐵 ∈ Cℋ
Assertion osumi ( 𝐴 ⊆ ( ⊥ ‘ 𝐵 ) → ( 𝐴 +ℋ 𝐵 ) = ( 𝐴 ∨ℋ 𝐵 ) )

Proof

Step Hyp Ref Expression
1 osum.1 ⊢ 𝐴 ∈ Cℋ
2 osum.2 ⊢ 𝐵 ∈ Cℋ
3 1 a1i ⊢ ( 𝐴 ⊆ ( ⊥ ‘ 𝐵 ) → 𝐴 ∈ Cℋ )
4 2 a1i ⊢ ( 𝐴 ⊆ ( ⊥ ‘ 𝐵 ) → 𝐵 ∈ Cℋ )
5 1 2 chsscon2i ⊢ ( 𝐴 ⊆ ( ⊥ ‘ 𝐵 ) ↔ 𝐵 ⊆ ( ⊥ ‘ 𝐴 ) )
6 5 biimpi ⊢ ( 𝐴 ⊆ ( ⊥ ‘ 𝐵 ) → 𝐵 ⊆ ( ⊥ ‘ 𝐴 ) )
7 3 4 6 chscl ⊢ ( 𝐴 ⊆ ( ⊥ ‘ 𝐵 ) → ( 𝐴 +ℋ 𝐵 ) ∈ Cℋ )
8 1 chshii ⊢ 𝐴 ∈ Sℋ
9 2 chshii ⊢ 𝐵 ∈ Sℋ
10 8 9 shjshseli ⊢ ( ( 𝐴 +ℋ 𝐵 ) ∈ Cℋ ↔ ( 𝐴 +ℋ 𝐵 ) = ( 𝐴 ∨ℋ 𝐵 ) )
11 7 10 sylib ⊢ ( 𝐴 ⊆ ( ⊥ ‘ 𝐵 ) → ( 𝐴 +ℋ 𝐵 ) = ( 𝐴 ∨ℋ 𝐵 ) )