# 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).

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]