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	$⊢ A \in C_{ℋ}$
	osum.2	$⊢ B \in C_{ℋ}$
Assertion	osumi	$⊢ A \subseteq ⊥ (B) \to A +_{ℋ} B = A \lor_{ℋ} B$

Proof

Step	Hyp	Ref	Expression
1		osum.1	$⊢ A \in C_{ℋ}$
2		osum.2	$⊢ B \in C_{ℋ}$
3	1	a1i	$⊢ A \subseteq ⊥ (B) \to A \in C_{ℋ}$
4	2	a1i	$⊢ A \subseteq ⊥ (B) \to B \in C_{ℋ}$
5	1 2	chsscon2i	$⊢ A \subseteq ⊥ (B) \leftrightarrow B \subseteq ⊥ (A)$
6	5	biimpi	$⊢ A \subseteq ⊥ (B) \to B \subseteq ⊥ (A)$
7	3 4 6	chscl	$⊢ A \subseteq ⊥ (B) \to A +_{ℋ} B \in C_{ℋ}$
8	1	chshii	$⊢ A \in S_{ℋ}$
9	2	chshii	$⊢ B \in S_{ℋ}$
10	8 9	shjshseli	$⊢ A +_{ℋ} B \in C_{ℋ} \leftrightarrow A +_{ℋ} B = A \lor_{ℋ} B$
11	7 10	sylib	$⊢ A \subseteq ⊥ (B) \to A +_{ℋ} B = A \lor_{ℋ} B$

Metamath Proof Explorer

Theorem osumi

Proof