# Greatest lower bound of Vector Space

## Homework Statement

Prove:
The set S(V) of all subspaces of a vector space V is a complete lattice under set inclusion, with smallest element {0}, largest element V, meet
$$glb(S_{i} | i \in K) = \cap_{i \in K} S_{i}$$
and join
$$lub(S_{i} | i \in K) = \sum_{i \in K} S_{i}$$

(Btw, how can I write underneath a symbol instead of at the subscript position?)

Solution:
I know there exists a glb and a lub since this is a complete lattice.

I'll start with the lub (least upper bound) and I'll try to show first that if S,T are subspaces of V then:
$$S + T = lub(S,T)$$

Let S,T be subspaces of V. Then there exist vectors $$s \in S$$ and $$t \in T$$.

Since $$s,t \in V$$(since S and T are subspaces of V) then $$s + t \in V$$.
Since all vectors in S,T are in V (I think I'm being repetitive) then all vectors from S summed with all vectors from T is in V hence: S + T is a subspace of V.

Should be obvious that S + T is least upper bound of S and T (is this obvious)?

Since S + T is a subspace of V call it U. Let W be a subspace of V then:
W + U = (S + T) + U = lub{U,V}
You can repeat this to finally show that:
$$lub{S_{i} | i \in K} = \sum_{i \in K} S_{i}$$

I think I can show the GLB in the same way.

I know this isn't very robust and probably longer than what is needed (I'm only starting out writing proofs).

Related Calculus and Beyond Homework Help News on Phys.org
All correct, but there are two remarks I'd like to make:

Should be obvious that S + T is least upper bound of S and T (is this obvious)?
No, I'm sorry, this is not obvious to me.

Since S + T is a subspace of V call it U. Let W be a subspace of V then:
W + U = (S + T) + U = lub{U,V}
You can repeat this to finally show that:
$$lub{S_{i} | i \in K} = \sum_{i \in K} S_{i}$$
You say that you repeat this argument, but if you do so, you will only obtain the result for finitely many subspaces. I.e. you will obtain that

$$lub\{U_1,...,U_n\}=U_1+...+U_n$$

But I is infinite here. So this case must still be dealt with...
In fact, if you find the result for two factors S and T, then the argument will generalize to infinitely many factors. So focus on the result for two factors for now...

I think I can show the GLB in the same way.

I know this isn't very robust and probably longer than what is needed (I'm only starting out writing proofs).[/QUOTE]