Does a complete Lexicographical Ordering implies component-wise completeness of Orders (and vice versa)?

Question

Question

Let $X$ be the product of $n$ sets $X_i$ each of which is ordered by $\succsim^L$. Show that the lexicographic order $\succsim^L$ is complete if and only if the component orders $\succsim_i$ are complete.

Note: All $X_i$ are ordered by $\succsim_i$

Source: FOUNDATIONS OF MATHEMATICAL ECONOMICS, Michael Carter, Exercise 1.35

For the $\Rightarrow$ part, I attempted a direct proof (by induction) but I couldn't bridge the gap in the inductive step in a succinct manner. Also I tried to tackle this proposition by proving its contra-positive which led me to nowhere. For the $\Leftarrow$ part, I think again an induction is necessary but again I couldn't close out the argument.

Answer 1

If $X = \varnothing$ it is direct let us consider that this is not the case.

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So let us do the implication first.

We suppose that $\leq_L$ is complete. Let us take $k \leq n$ and $x,y \in X_k$. We just take $\forall i \neq k, \, x_i \in X_i$.
Since $\leq_L$ is complete, we have either $$(x_1,\dots,x_{k-1},x,x_{k+1},\dots,x_n) \leq_L (x_1,\dots,x_{k-1},y,x_{k+1},\dots,x_n)$$ or $$(x_1,\dots,x_{k-1},y,x_{k+1},\dots,x_n) \leq_L (x_1,\dots,x_{k-1},x,x_{k+1},\dots,x_n)$$ Since the first $k-1$ terms are the same, we have either $x \leq_k y$ or $y \leq_k x$ which means that $\leq_k$ is complete.

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Now for the return, if all the $\leq_k$ are complete, let us suppose that there are $x=(x_i)_i$ and $y = (y_i)_i$ such that none of the following assertion are true:

• $x \leq_L y$
• $y \leq_L x$

That means that $x_1 = y_1$, otherwise we would have $x <_L y$ or $y <_L x$, and we can iterate (or you can make it an induction if you wish) with the same reasoning and get that $x_2 = y_2$, etc.
That means that $\forall k\leq n, \, x_k = y_k$, i.e. $x=y$ which is absurd. Hence $\leq_L$ is complete.