Is this relation $R : (f,g) \in R$ on the set $\{f:\mathbb{R} \rightarrow \mathbb{R} \}$ partial order relation? And total order relation?

Question

R is defined aas the relation on the set of all functions from $\mathbb{R}$ to $\mathbb{R}$.

$$ R: (f,g) \in R \iff f(x) \leq g(x), x \in \mathbb{R}$$

The question is, is R a relation of partial order? If yes, is it a relation of total order?

My answer:

It's reflexive since: $$(f,f) \in R \iff f(x) \leq f(x)$$ It's transitive: $$(f,g) \in R, (g,z) \in R \hookrightarrow (f,z) \in R$$

$$\iff f(x) \leq g(x), g(x) \leq z(x) \hookrightarrow f(x) \leq z(x)$$

It's antisymmetric:

$$(f,g) \in R, (g,f) \in R \hookrightarrow f=g$$ $$\iff f(x) \leq g(x), g(x) \leq f(x) \hookrightarrow f(x)=g(x)$$

Therefore, it's a relation of partial order.

Then the second part. For it to be a relation of total order it has to be a relation of partial order and also satisfy this condition:

$$\forall f,g \hookrightarrow \begin{cases}\text{either: } (f,g) \in R\\ \text{either: }(g,f) \in R \end{cases}$$

$$\iff \begin{cases}\text{either: } f(x) \leq g(x)\\ \text{either: }g(x) \leq f(x) \end{cases}$$

This is the part where I have doubts, because maybe a there are two functions $f$ and $g$, such that the intersection or their domains $D_f \cap D_g = \emptyset$, then they wouldn't be comparable by R. So is it not a relation of total order?

Thank you so much for your help!

(As a side note: does anyone know how to 'exclude' some text from latex, inside a latex construction, for example, where I wrote 'either', how could I make it so that it's not latex?)

"such that the intersection or their domains $D_f\cap D_g=\emptyset$"...You already specify that $f: \mathbb{R}\to \mathbb{R}$. Therefore, $D_f=\mathbb{R}$ by definition — Adam Martens, Dec 15 '19 at 18:01

score 2 · Answer 1 · answered Dec 15 '19 at 18:03

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This is not a total ordering. Consider $f=1_{[0,\infty)}$ and $g=1_{(-\infty,0)}$. Then is is neither the case that $(f,g)\in R$ nor $(g,f)\in R$.

answered Dec 15 '19 at 18:03

Adam Martens

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What does ${[0,\infty)}$ and ${(-\infty,0)}$ mean? – Belen Dec 15 '19 at 18:20
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These are just intervals on the real line. The notation $1_{A}$ is the indicator function of the set $A$. That is $1_A(x)=1$ if $x\in A$ and $1_A(x)=0$ if $x\not\in A$. – Adam Martens Dec 15 '19 at 18:21
So, in this case, we just can't choose an $x$, such that $\exists f(x)$ and $\exists g(x)$ at the same time, right? – Belen Dec 15 '19 at 18:23
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Right. You need $f$ and $g$ that work for all $x\in \mathbb{R}$ simultaneously – Adam Martens Dec 15 '19 at 18:24

score 2 · Accepted Answer · answered Dec 15 '19 at 18:04

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This is the part where I have doubts, because maybe a there are two functions $f$ and $g$, such that the intersection or their domains $D_f \cap D_g = \emptyset$, then they wouldn't be comparable by R. So is it not a relation of total order?

We're talking about functions from $\mathbb{R}$ to $\mathbb{R}$, so their domains are $\mathbb{R}$ in all cases. So, that's not something to worry about.

However, consider $f(x)=x$ and $g(x)=2x$. Do you see that we have neither $fRg$ nor $gRf$?

answered Dec 15 '19 at 18:04

Bram28

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Why not? We have that: $f(x)=x \leq 2x=g(x)$ – Belen Dec 15 '19 at 18:16
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Not if $x<0$. Then $f>g$ – Adam Martens Dec 15 '19 at 18:22
I see. Does this break the partial order or only the total order? – Belen Dec 15 '19 at 18:25
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@SofiaB.Lopez For $x \geq0$ we have $f(x) \leq g(x)$, but for $x<0$ we hjave $g(x) < f(x)$, and so we don;t have that $f(x) \leq g(x)$ for all $x \in \mathbb{R}$ – Bram28 Dec 15 '19 at 18:26
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@SofiaB.Lopez It breaks total order. Your proof that $R$ is a partial order is correct. – Bram28 Dec 15 '19 at 18:27

Is this relation $R : (f,g) \in R$ on the set $\{f:\mathbb{R} \rightarrow \mathbb{R} \}$ partial order relation? And total order relation?

2 Answers2