Lattice of continuous functions Let $\mathcal C
[0
,
1]$ be the set of continuous functions from $[0
,
1]$ to the reals. Define
$\leq$
on
$\mathcal C
[0
,
1]$ by
$f
\leq
g \iff
f
(
a
)
\leq
g
(
a
)\; \forall
a
\in
[0
,
1]
$.
Show that
$\leq$
is a partial order which
makes
$\mathcal C
[0
,
1]$ into a lattice
 A: To show that a partially ordered set $\langle \mathbb{P},\preceq\rangle$ is a lattice, you need to show that any two elements of $\mathbb{P}$ have both a least upper bound and a greatest lower bound. 
In the case of $\langle \mathcal{C}[0,1],\leq\rangle$, this means that for $f$ and $g$, you must find continuous functions $f\vee g$ and $f\wedge g$ with the following properties:


*

*$f\leq f\vee g$

*$g\leq f\vee g$

*$\forall h\in \mathcal{C}[0,1] \big(\big((f\leq h) \& (g\leq h)\big) \to (f\vee g)\leq h \big)$ 

*$f\geq f\wedge g$

*$g\geq f\wedge g$

*$\forall h\in \mathcal{C}[0,1] \big(\big((f\geq h) \& (g\geq h)\big) \to (f\wedge g)\geq h \big)$ 


In fact, this is pretty straightforward.  The obvious choices for $f\vee g$ and $f\wedge g$ work, and the proofs of 1-6 are more or less immediate.  The only tricky thing is to show that  $f\vee g$ and $f\wedge g$ are actually continuous.

In fact, if you already know that $[0,1]$ is a lattice, and that the function space $$[0,1]^{[0,1]}=\prod_{i\in [0,1]}[0,1]$$ is ipso facto also a lattice, then you can take a shortcut.  The lattice $[0,1]^{[0,1]}$ is the set of all functions from $[0,1]$ to $[0,1]$ (not just continuous ones), and the $\vee$ and $\wedge$ functions there are defined pointwise (since it's a product).  
Then, to prove that $\mathcal{C}[0,1]$ is a sublattice of $[0,1]^{[0,1]}$, all you need to do is show that $\mathcal{C}[0,1]$ is closed under the operations $\wedge$ and $\vee$ inherited from $[0,1]^{[0,1]}$.  In other words, show that the (pointwise) meet and join of continuous functions are also continuous.
This is similar to proving that $H$ is a group by showing that $H$ is a subset of a group $G$, and that $H$ is closed under the group operations. 
A: Hint:
1.reflexivity:$f\leq f$ it's obvious
2.antisymmetry $f(a)\leq g(a)$ and $g(a)\leq f(a)$ for all $a\in [0, 1]$ then $f=g$
3.transitivity $f(a)\leq g(a)$ and $g(a)\leq h(a)$ then $f(a)\leq g(a)$
It's lattice because each $f\in C[0,1]$ is continuous and $[0,1]$ is compact so has sup and inf on this interval.
