Is infinite product of (complete) lattices a (complete) lattice? I was wondering if $(1)$ the product of an infinite number of lattices is a lattice and $(2)$ if the product of an infinite number of complete lattices is a complete lattice.

More precisely, suppose that $\{L_{a},a\in A\}$ is a family of lattices, where A is potentially infinity (e.g., $A=\mathbb{N}$ or $A=[0,1]$). Is $L=\times_{a\in A} L_{a}$ a lattice? What if  $\{L_{a},a\in A\}$ is a family of complete lattices. Is $L=\times_{a\in A} L_{a}$ a complete lattice as well? I know that both statements hold for finite products, but do they extend to infinite products?

I believe that the answer to the first question is yes. We simply define the "standard" order on $L$ where $x\geq y$ if $x_{a} \geq y_{a}$ for all $a \in A$, $x \vee y = \{x_a \vee y_a \}_{a \in A}$, and $x \wedge y = \{x_a \wedge y_a \}_{a \in A}$. 
The second question seems to be more complicated. I am also inclined to say that it is true, but I am less sure of it. In particular, I was thinking if the approach takes from finite products could be extended here. In particular, let $K\subset L$. Define $K_a=\{x_a\in L_a | \exists y\text{ in } K \text{ with } y_a=x_a  \}$. Then $K_a$ is a subset of $L_a$ and since $L_a$ is complete then $sup(K_a)$ exists in $L_a$. Then $sup(K)=\{sup(K_a)\}_{a\in A}$ which belongs to $L$ since $sup(K_a)\in L_a$ for all $a \in A$.

 A: Your $K_a$ is called the $a$-th projection of $K$, and you want to prove that
$$\bigvee K = \left(\bigvee K_a:a \in A\right).$$
(Notice that you notation, is not correct.
When you write $x \vee y = \{x_a \vee y_a\}_{a \in A}$, you saying that the join of $x$ and $y$ is given by that set, but it must be a sequence, because the order matters.
So it should be $x \vee y = (x_a \vee y_a)_{a \in A}$ or $x \vee y = (x_a \vee y_a :a \in A)$, or something similar.)
To prove the above equality, let $v_a = \bigvee K_a$, and $v = (v_a:a \in A)$.
Now we prove that $v$ is the join of $K$.
First, let $k \in K$.
For each $a \in A$, we have that $k_a \in K_a$, whence $k_a \leq v_a$, yielding $k \leq v$.
Now let $u = (u_a:a\in A)$ be any upper bound of $K$, that is, $k \leq u$ for all $k \in K$.
It follows that $k_a \leq u_a$, for every $k \in K$ and every $a \in A$, whence, by the definition of supremum, $v_a \leq u_a$, for every $a \in A$, and therefore $v \leq u$.
So $v$ is the least upper bound of $K$, QED.
