Orthocomplement Reverses Subset

Theorem

Let $\struct {V, \innerprod \cdot \cdot}$ be an inner product space space.

Let $A, B$ be subsets of $V$ with $A \subseteq B$.

Then:

$B^\perp \subseteq A^\perp$

where $\perp$ denotes orthocomplementation.

Proof

Let:

$h \in B^\perp$

Then, from the definition of orthocomplement, we have:

$h \perp b$ for each $b \in B$.

Since $A \subseteq B$, we in particular have:

$h \perp a$ for each $a \in A$.

So, from the definition of orthocomplement:

$h \in A^\perp$

So:

$h \in B^\perp$ implies $h \in A^\perp$.

So, from the definition of subset:

$B^\perp \subseteq A^\perp$

$\blacksquare$