We have preference relation $\succeq $ on $X = \mathbb{R}^n$ represented by u(x) = $\sum_{i=1}^{n} x_i$. Show that $\succeq $ is rational, continuous and strong monotonic.

So I think I have rationality. Since u(x) and u(y) will be real numbers, completeness and transitivity of the reals can be used to work backwards that $\succeq $ is complete and transitive as well.

For continuity, Id like to show that the upper and lower contour sets are closed but Im not sure how to go about it with the utility representation.

For strong monotonicity, can I say that if x > y the $\sum_{i=1}^{n} x_i$ will be greater than the $\sum_{i=1}^{n} y_i$ for all x and y which implies that u(x) > u(y) and $\succeq $ is strong monotonic?


1 Answer 1


Your thought process is correct in proving rationality and strict monotonicity. As for continuity of preference relations, remember the following equivalence: $\forall x\in X, U(\succsim,x) \ \text{and} \ L(\succsim,x)$ are closed on $X \ \Leftrightarrow \ \succsim$ is a closed subset of $X \times X$.

Additionally, $\succsim$ is a closed subset of $X \times X \ \Leftrightarrow \ $ if for any two convergent sequences $\left\{x^n\right\} \rightarrow x$ and $\left\{y^n\right\} \rightarrow y \ $ such that $x^n \succsim y^n \ \forall n\in \mathbb{N}$, then $x\succsim y$.

Prove that if a preference relation $\succsim$ on $X\equiv \mathbb{R}^n$ is represented by a function $u(x)=\sum_{i=1}^n x_i$, then $\succsim$ is rational, continuous and strictly (strong) monotonic.

Note that $\succsim$ is represented by $u(x)$ implies that $\forall x,y\in $X$, x\succsim y \Leftrightarrow u(x)≥u(y)$.


The Axiom of Trichotomy states that $\forall a,b\in \mathbb{R}, a≥b \ \text{or} \ b≥a$. Thus, $\forall u(x),u(y)\in \mathbb{R}, u(x)≥u(y) \ \text{or} \ u(y)≥u(x) \Rightarrow x\succsim y \ \text{or} \ y \succsim x, \ \forall x,y\in X \Rightarrow \ \succsim \ $ is complete.

Since $\succsim$ is complete, i.e. $\forall x,y\in X, x\succsim y$ or $y\succsim x$, it holds that $u(x)≥u(y)$ or $u(y)≥u(x), \forall x,y\in X$.


Let $u(x)≥u(y)$ and $u(y)≥u(z)$. By the transitive property of real numbers, $u(x)≥u(z)$. In terms of the preference relation, $x \succsim y$ and $y \succsim z \Rightarrow \ x \succsim z \Rightarrow \ \succsim \ $ is transitive.


Let $\left \{ x^t \right \} \rightarrow x$ and $\left \{y^t\right \}\rightarrow y$ be arbirary convergent sequences on $X$, where $x^t \succsim y^t \ \forall t\in \mathbb{N}$. This implies that $u(x^t)≥u(y^t) \ \forall t\in \mathbb{N} \Rightarrow \ u(x)≥u(y) \ $ by the Order Limit Theorem $\Rightarrow \ $ Continuity of $\succsim $

Strictly Monotone

Let $x >> y$, which implies that $\forall i=(1,...,n), \ x_i > y_i $. We must show that $x \succ y$.

$x >> y \Rightarrow \sum_{=1}^n x_i > \sum_{i=1}^n y_i \Rightarrow u(x) > u(y) \Rightarrow x \succ y$. $\square$


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