Continuous rational and monotone preference relation implies $x\succsim0$?

Question

I updated my proof to a general version as follows: please share your thoughts & 2cent. Thanks

Show a monotone continuous complete preorder on $\mathbb{R}^L_+$ has $y\geq x\rightarrow y\succsim x$.

Point of Clarification
$X=\mathbb{R}^L_+$
Preorder means the usual reflexivity and transitivity.
Complete means for any $x,y\in X$, have $x\succsim y$ or $y\succsim x$
Continuous means the relation is preserved under limits.
Monotone means for any $x,y\in X$, if $y\gg x$, then $y\succ x$.
$\succ$ and $\sim$ are respectively asymmetic and symmetric parts of $\succsim$

Outline of Proof
Go through two cases: (1) $y\gg x$. Easily get the result by definition. (2) Some components are equal while else y is strictly greater x. Use continuity where you add a sequence of small positive $\epsilon$ to y, making it a sequence $y^n_\epsilon$ where for every n, $y^n_\epsilon\gg x$ $\forall n$.

Proof
Suppose $\succsim$ is a monotone, continuous, complete preorder on $X=\mathbb{R}^L_+$.
Case (1) $y\gg x$ (i.e. $y_i>x_i$ $\forall i\in B=\{1,\dots,L\}$).
By definition, $y\succ x$, which implies $y\succsim x$.

Case (2) $y_j=x_j$ for some $j\in B$. For $\forall k\not=j,k\in B, y_k>x_k.$
For some $\epsilon>0$, let the sequence $\epsilon^n\in\mathbb{R}^L_+$ such that $\epsilon_j=\epsilon$, $\epsilon_k=0$.
Denote the sequence $y^n_\epsilon=y+\epsilon^n$.
Then, for any $\epsilon>0$ and $\forall n$, $y^n_\epsilon\succ x$, hence $y^n_\epsilon\succsim x$.
By continuity of $\succsim$, $$\lim_{\epsilon \to 0} {y^n_\epsilon} = y$$
Hence, $y\succsim x$. $\blacksquare$

OLDER VERSION

My question is what is the valid reasoning behind that continuous rational and monotone preference relation implies $x\succsim0$. I have put a proof below and would appreciate if you share your 2 cent on the validity/rigor of the proof. Thanks!

Suppose $x\in\mathbb{R}^L_+=\{x\in\mathbb{R}^L:x_l\geq0$ $\forall l=1,\dots,L\}$.

Claim: For every $x\in\mathbb{R}^L_+$, monotonicity implies $x\succsim0$.

Proof:

(1) Suppose $x=(0,\dots,0)$. Then, $x\sim0$ is possible.

(2) Suppose $x\gg y$. Then, by Definition of monotone preference, $x\succ y$ is possible.

(3) Suppose $\exists$ some $j$ such that $x_j>0$ and $1\leq j\leq L$.Then, I have the following process of elimination:

• $x\succsim0$ is possible.
• $x\succsim0$ and $0\succsim x$ $\iff x\sim0$ is possible.
• $x\succsim0$ but not $0\succsim x$ $\iff x\succ0$ is possible.
• $0\succsim x$ but not $x\succsim 0$ $\iff 0\succ x$ is impossible.

The single preference relation that is common in all three scenarios is $x\succsim0$. Hence, for every $x\in\mathbb{R}^L_+$, monotonicity implies $x\succsim0$. Q.E.D

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Reference: Proposition 3.C.1 in Microeconomic Theory by Andreu Mas-Colell, Michael Whinston, and Jerry Green.

Answer 1

If we take the definition of monotonicity to be if $x\geqq y$ then $x \succeq y$, you can simplify the proof (though it looks right).

Note $\mathbf{0}\leq x$ for all $x\in \mathbb{R}_+^l$. So by the definition of monotonicity (essentially replacing $y$ with $\mathbf{0}$ above), $x\succeq \mathbf{0}$. I don't think continuity is required (check lexicographic preferences to see it is not necessary for the stated result).

Answer 2

I did not understand the last part when you said $0≻x$ is impossible. I think it's necessary to use continuity + monotonicity in (3).

For instance, let's take the $\Bbb R^2$ case. Assume we have two elements $x_1=(0,1)$ and $x_2=(1,0)$. Then, any combination is greater than $0$: $\alpha*x_1+(1-\alpha)*x_2 >>0$ if $\alpha \in (0,1)$. Re-definining $\alpha=1/n$. Then, we have built a sequence $x_n=(1/n)*x_1+(1-1/n)*x_2$ such that $x_n >> 0$ for any $n=2,3,...$. By monotonicity, $x_n≿0$ for all $n=2,...$. Finally, by continuity of preferences, $\lim_{n\to\infty}x_n≿0$. That is, $x_2≿0$ (the same for $x_1≿0$).

This can be done in $\Bbb R^n$ and for any vector that have $x_j=0$ for some $j$. Hence, adding this with the proof you have made for the other cases ($x=0$ and $x>>0$) we have $x≿0$. The (1) case is implied by completeness, which imply reflexivity.