I need to prove the following result:
For a monotone, continuous, complete, and transitive preference relation $\succsim$ on $X=\mathbb{R}^L_+$, $y\geq x$ implies $y\succsim x$.
I tried it myself, but I'm not sure whether it is correct. May I ask someone to check my work? Thank you so much!
Let $\succsim$ be a monotone, continuous, complete, and transitive preference relation on $X=\mathbb{R}^L_+$. Suppose first that $y\gg x$; that is, $y_i>x_i$ for all $i=1,\dots,L$. Then by the definition of monotonicity, $y\succ x$ and thus $y\succsim x$.
Now suppose that $y_i=x_i$ for some $i\in\{1,\dots,L\}$ and $y_k>x_k$ for all $k\neq i$. (So $y\geq x$.) Let $\{y^n\}$ be a sequence of points in $\mathbb{R}^L_+$ such that $y_k^n=y_k$ for $k\neq i$ and $y_i^n=y_i+\frac{1}{n}$. Let $\{x^n\}$ be a sequence of points in $\mathbb{R}^L_+$ such that $x_k^n=x_k$ for $k\neq i$ and $x^n_i=x_i+\frac{1}{n+1}$. Then for each $n$, $y^n_k = y_k > x_k = x^n_k$ and $y^n_i = y_i + \frac{1}{n} = x_i + \frac{1}{n} > x_i + \frac{1}{n+1} = x^n_i$. Thus $y^n\gg x^n$, and monotonicity implies $y^n\succ x^n$, and so $y^n\succsim x^n$. So we have constructed a sequence of pairs $\{(x^n,y^n)\}$ such that $y^n\succsim x^n$ for all $n$, $x=\lim_{n\to\infty}x^n$, and $y=\lim_{n\to\infty}y^n$. Then continuity implies $y\succsim x$.
Some Background Information:
Notation$\quad$ $\mathbb{R}^L_+ = \left\{x\in\mathbb{R}^L:x_l\geq0\ \text{for}\ l=1,\dots,L\right\}$.
Notation$\quad$ Let $x,y\in\mathbb{R}^N$ so $x=(x_1,\dots,x_N)$ and $y=(y_1,\dots,y_N)$. Then
(i) $x\geq y$ means $x_n\geq y_n$ for all $n=1,\dots,N$;
(ii) $x\gg y$ means $x_n>y_n$ for all $n=1,\dots,N$.
Definition$\quad$ The preference relation $\succsim$ on $X$ is rational if it possesses the following two properties:
(i) Completeness. For all $x,y\in X$, we have $x\succsim y$ or $y\succsim x$ (or both).
(ii) Transitivity. For all $x,y,z\in X$, if $x\succsim y$ and $y\succsim z$, then $x\succsim z$.
Definition$\quad$ The preference relation $\succsim$ on $X$ is monotone if $x\in X$ and $y\gg x$ implies $y\succ x$.
Definition$\quad$ The preference relation $\succsim$ on $X$ is continuous if it is preserved under limits. That is, for any sequence of pairs $\{(x^n,y^n)\}_{n=1}^{\infty}$ with $x^n\succsim y^n$ for all $n$, $x=\lim_{n\to\infty}x^n$, and $y\lim_{n\to\infty}y^n$, we have $x\succsim y$.
