Lattice of continuous functions

Question

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

Answer 1

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:

1. $f\leq f\vee g$
2. $g\leq f\vee g$
3. $\forall h\in \mathcal{C}[0,1] \big(\big((f\leq h) \& (g\leq h)\big) \to (f\vee g)\leq h \big)$
4. $f\geq f\wedge g$
5. $g\geq f\wedge g$
6. $\forall h\in \mathcal{C}[0,1] \big(\big((f\geq h) \& (g\geq h)\big) \to (f\wedge g)\geq h \big)$

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.

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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.

Answer 2

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.