For the first question: yes we can show in graph.

Second Stefan is right because lexicographic prefences are not continous.

If there are two goods x,y and U(x1,y1)>U(x2,y2) if x1>x2 or if x1=x2 and y1>y2 then can't we represent this as region. Any point east of x1,y1 or due north is preferred. There is no utility curve, however, since any other point is either preferred or not preferred. Not sure how useful the graph is though.

The indifference sets contain only one element, i.e. given any bundle all the other bundles are either more desirable or less desirable. You can of course visualize this on a 2-good quadrant (not very informative, as John suggests). If the choice set is finite, i.e. contains a finite number of bundles, then - provided the preferences are transitive and complete - you can always find a numerical representation of preferences, even lexicographic ones: this might be relevant in empirical applications. You need continuity of preferences when the choice set is not finite. The proof that lexicographic preferences cannot be represented by a utility function (whether continuous or not continuous) might go as follows.

First, you cannot establish a one-to-one correspondence between the real numbers and the rational numbers. Clearly there are many real numbers that are not rational (for example the square root of 2 is not rational).

Second, in any interval of real numbers there is at least one rational number.

Now, keeping the above in mind, let’s go back to lexicographic preferences on two-dimension bundles. We want to show that we can’t have a utility function.

Consider (x1, 2) and (x1, 1) and suppose instead we have a utility function that represents the preferences.

A utility function would produce two real numbers such that u(x1, 2) > u(x1, 1).

In the interval between these two real numbers there exists a rational number, call it r( x1).

Now consider (x'1, 2) and (x’1, 1) with x’1 > x1 .

Given lexicographic preferences, both u(x’1, 2) and u(x’1, 1) must be larger than u(x1, 2): therefore for any rational number r(x’1) in the interval [u(x’1, 2), u(x’1, 1)] we must have r(x’1) > r(x1). But then we would have a one-to-one correspondence between the real numbers (x1, x’1…) and the rational numbers (r(x1), r(x’1) ….): which is impossible!

Very good explanation Mr. Colombino. I'll use it in my classes.

Thanks Ugo, that makes sense, though note that there being real numbers that are not rational does not prevent the existence of a one-to-one correspondence (there are reals that are not irrational, but the two sets have the same cardinality).

This same thought held on to me after I first learnt about Lexicographic preferences. The best way was to try it and go to sleep.

The answers given above are correct. So I will not add.

However, if the goods are discrete (with whatever small FINITE size), then this is what you are looking for: