Given a vector space V, the set of all subspaces is a lattice

Question

Partial order considered in this question is set inclusion. How do I complete part (1a) and (2b)? I suppose that once solved one of the two the other one will use the same trick.

Lemma. Given a vector space $V$, the set $\mathcal{S}(V):=\{W|W\text{ subspace of }V\}$ is a lattice.

Proof. Given $W_1,W_2\in \mathcal{S}(V)$, we prove (1) $\text{lub}\{W_1,W_2\}=W_1+W_2$; and (2) $\text{glb}\{W_1,W_2\}=W_1\cap W_2$.

(1): By Lemma \ref{lem_ss}, $W_1+ W_2$ is a subspace. (1a) $\text{lub}\{W_1,W_2\}\subset W_1+W_2$: By contradiction. Suppose there is a vector $v\in \text{lub}\{W_1,W_2\}$ and $v\notin W_1+W_2$. [???]

(1b) $W_1+W_2\subset \text{lub}\{W_1,W_2\}$: A vector $v\in W_1+W_2$ can be expressed as $v=v_1+v_2$, $v_1\in W_1$ and $v_2\in W_2$. Since $\text{lub}\{W_1,W_2\}$ is vector space containing $W_1$ and $W_2$, it contains also the sum of their elements.

(2): By Lemma \ref{lem_ss}, $W_1\cap W_2$ is a subspace. (2a): $\text{glb}\{W_1,W_2\}\subset W_1\cap W_2$: From being a lower bound follows that for any $v\in \text{glb}\{W_1,W_2\}\subset W_1$, $v\in W_1$, analogously $v\in W_2$, and therefore $v\in W_1\cap W_2$.

(2b): $W_1\cap W_2\subset \text{glb}\{W_1,W_2\}$ [???]

Answer 1

I think you're getting turned around a bit by starting from the assumption that $lub\{W_1, W_2\}$ and $glb\{W_1, W_2\}$ already exist, and then comparing them with the spaces $W_1+W_2$ and $W_1\cap W_2$. Instead, just argue directly that each of the spaces in question satisfies the relevant definition. For example, to show that $W_1+W_2$ is the least upper bound of $W_1$ and $W_2$, you need to show:

• $W_1+W_2$ is a subspace containing $W_1$ and $W_2$. That is, $W_1+W_2$ is an upper bound of $W_1$ and $W_2$.

• If $W$ is a subspace containing both $W_1$ and $W_2$, then $W$ contains $W_1+W_2$. That is, $W_1+W_2$ is the least upper bound of $W_1$ and $W_2$.

E.g. for the second bulletpoint above, we can argue as follows. Suppose $W$ is some upper bound of $W_1$ and $W_2$. Fix $v\in W_1+W_2$; we want to show that $v\in W$. Well, since $v=W_1+W_2$ we can find vectors $w_1\in W_1, w_2\in W_2$ such that $v=w_1+w_2$. Since $W$ contains $W_1$, we have $w_1\in W$; similarly, $w_2\in W$. But since $W$ is a subspace, it's closed under addition, so $w_1+w_2=v$ is in $W$. Since $v$ was an arbitrary element of $W_1+W_2$, what we've really shown is that every element of $W_1+W_2$ is contained in $W$ - that is, $W\supseteq W_1+W_2$.

Part 2 is proved similarly.