What Are the Orthogonal Complements of Union and Intersection in Hilbert Spaces?

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Let M and N be two subsets of a hilbert space H.
What are orthogonal complements of following sets:
1) The union of M and N.
2) The intersection of M and N.
 
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So, what did you try already??
 
x\in (M\cap N)^\bot means
<x,y>=0 for all y\in M \text{ and } y\in N and hence

x\in M^\bot\cap N^\bot.. The other direction is obvious, so we get (M\cap N)^\bot=M^\bot\cap N^\bot
This is for the intersection, but I strongly think that I had a mistake.
 
x\in (M\cap N)^\bot means
<x,y>=0 for all y\in M \text{ and } y\in N and hence

x\in M^\bot\cap N^\bot.. The other direction is obvious, so we get (M\cap N)^\bot=M^\bot\cap N^\bot
This is for the intersection, but I strongly think that I had a mistake.
 
I don't think the other inclusion is obvious.

In fact, I suspect (M\cap N)^\bot = M^\bot + N^\bot...
 
I do not know what do you mean by the sum of tow sets? and how can I prove that?
 
one can't, in general, sum two sets, but if we already have an addition defined, then:

A + B = {a+b : a is in A, b is in B}.
 
How did you guess that it is the sum of?

since my goal is to reach this equality.

How can we prove it?
 
normally, to prove 2 sets are equal, you show they contain each other.
 
  • #10
Of course, but I did not manege tp solve it!
On the other hand, (M\cap N)^\bot=M^\bot+N^\bot means M^\bot+ N^\bot\subset M^\bot\cap N^\bot, but this seems senseless, does not it?
 
  • #11
LikeMath said:
x\in (M\cap N)^\bot means
<x,y>=0 for all y\in M \text{ and } y\in N and hence

x\in M^\bot\cap N^\bot.. The other direction is obvious, so we get (M\cap N)^\bot=M^\bot\cap N^\bot
This is for the intersection, but I strongly think that I had a mistake.

your definition of (M\cap N)^\bot isn't correct here.

we don't know that <x,y> = 0 for all y in M and y in N, just those y that are in the intersection. if y is in M-N, all bets are off.
 
  • #12
Oh yes, that's the problem. But the problem now how can we prove that
(M\cap N)^\bot=M^\bot+N^\bot
 
  • #13
LikeMath said:
(M\cap N)^\bot=M^\bot+N^\bot

I think we should have some contraintes on M and N, do not I?
 
Last edited:
  • #14
LikeMath said:
I think we should have some contraintes on M and N, do not I?

Yeah, M and N should both be subspaces.
 
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