Lexicographic preference relation cannot be represented by a utility function I am stuck on the following exercise, related to preference relations and von-Neumann-Morgenstern utility function.

A farmer wants to dig a well in a square field $[0,1000]\times[0,1000]$. The preferences of the farmer on the possible locations are lexicographic, i.e:

If $x_1&lt;x_2$ then $(x_1,y_1)\prec(x_2,y_2)$ for all $y_1,y_2$.
If $x_1=x_2=x$, then $(x,y_1)\prec(x,y_2)$ iff $y_1 &lt; y_2$.

Initially, assume that the well location must have integer coordinates.
Is there a preference relation on lotteries, that satisfies the von-Neumann-Morgenstern axioms, and extends the lexicographic preference relation? If so, what is a linear utility function that represents this relation?

I think the answer is yes, and a possible linear utility function is: $u(x,y)=100000x + y$.

Now, assume that the well location can have real coordinates. Prove that there is no linear utility function that represents the preference relation on lotteries.
Which one of von-Neumann-Morgenstern axioms is violated by the preference relation on lotteries?

Here I am stuck. I don't understand why the utility function I suggested above doesn't work? And what axiom is violated here?
We can say more generally that lexical preferences are not representable using a continuous utility function. Lexical preferences are not continuous. Note the definition of a continuous preference relation.

The preference relation $\succeq$ is continuous if for any sequences of consumption bundles $(x_{i})_{i \in \mathbb{N}}$ and $(y_{i})_{i \in \mathbb{N}}$ with $x_{i} \to x$, $y_{i} \to y$, and $x_{i} \succeq y_{i}$ for each $i \in \mathbb{N}$, then $x \succeq y$. That is, continuity preserves the relation at the limit point.

Consider $(x_{i})_{i \in \mathbb{N}}$ defined by $x_{i} = (\frac{1}{2^{i}}, 0)$ and $(y_{i})_{i \in \mathbb{N}}$ defined by $y_{i} = (0, 1)$. Clearly, $x_{i} \succeq y_{i}$ for each $i \in \mathbb{N}$. However, $x_{i} \to (0, 0)$ while $y_{i} \to (0, 1)$. So this preference relation is not preserved at the limit point.

Even more generally, no utility function represents the lexical preference relation. I prove for $\mathbb{R}_{+}^{2}$, but this argument extends to $\mathbb{R}_{+}^{n}$ by projecting into $\mathbb{R}_{+}^{2}$. 

Proof: Suppose to the contrary that some utility function $u : \mathbb{R}_{+}^{2} \to \mathbb{R}$ represents $\succ_{lex}$. We thus have $u(x, 1) &gt; u(x, 0)$, as $(x, 1) \succ (x, 0)$. We construct the interval $I(x) = [u(x, 0), u(x, 1)]$. Now for any two distinct $x, y \in \mathbb{R}_{+}$, $I(x) \cap I(y) = \emptyset$ as we have either $x &gt; y$ or $y &gt; x$ (so WLOG, we have $(x, 0) \succ (y, 1)$). 

Define $\mathbb{I} = \{ I(x) : x \in \mathbb{R}_{+} \}$, and let $\phi : \mathbb{R}_{+} \to \mathbb{I}$ be given by $\phi(x) = I(x)$. Observe that $\phi$ is an injection, as each $I(x), I(y)$ are disjoint for distinct $x, y$.

Note that $\mathbb{Q}$ is dense in $\mathbb{R}$. So there exists a rational number in each interval. Define $\tau : \mathbb{I} \to \mathbb{Q}$ such that $\tau(I(x))$ returns a rational number contained in $I(x)$. So $\tau$ is an injection. We have $\tau$ composed with $\phi$ an injection, which implies $|\mathbb{R}_{+}| \leq |\mathbb{Q}_{+}|$, a contradiction. QED.