I'm given the following question:
Show that any utility function on a finite set of alternatives attains maximum and minimum values by using Bolzano-Weierstrass Thm.
Well, if it were the case that we were given a continuous function on a finite domain (to R) , we could easily say that finite sets are closed and bounded, therefore, by Heine-Borel compact, so by Bolzano-Weierstrass, function has max. and min. However, in this case, how can I show that the corresponding utility function is continuous and its domain is compact? I'm open to any help.
If the set of alternatives $X$ with $N$ elements is finite, $u(X)$ is finite: it is just the image of the $N$ utility indices of the finite alternatives. Order the $N$ elements $u_1,...,u_N$ and take the largest of the $N$ numbers; the inverse image is the maximizer.
If you really want to go overboard, you do the following: any function on a finite set in a metric space is continuous. For each element $x_i \in X$, there is a distance $\delta_i = \min_{j} d(x_i,x_j)/2$. If $d(x_i,x')<\delta_i$, then the open ball around $x_i$ contains only $x_i$. So for all $\varepsilon>0$ and $d(x_i,x')<\delta_i$, $d(u(x_i),u(x'))<\varepsilon$ (namely, zero). So $u$ is continuous on $X$. A finite set of points is compact, since every open cover of $X$ has a finite subcover consisting of, at most, $N$ elements. Therefore $u$ is continuous on a compact set and has a maximizer and minimizer in the set by way of Weierstrass.