Continuous rational and monotone preference relation implies $x\succsim0$? I updated my proof to a general version as follows: please share your thoughts &amp; 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 $X=\mathbb{R^L_L}$.
   Case (1) $y\gg x$ (i.e. $y_i&gt;x_i$ $\forall i\in B=\{1,\dots,L\}$).
   By definition, $y\succ x$, which implies $y\succsim x$.
    Case (2) $y_j=x_j$ for some $j\in B$. For $\forall k\not=j,k\in B, y_k&gt;x_k.$
   For some $\epsilon&gt;0$, 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 $\epsilon&gt;0$ 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 $x_j&gt;0$ 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
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).