# Understanding the Continuity axiom of preference

Let $$x^{1}, x^{2}, \cdots \to x$$ where each $$x^{i}$$ and $$x$$ are elements of the set of consumption bundle or the choice set $$X$$. If $$x^{i} \succeq y$$ for each $$i \geq 1$$ then $$x \succeq y$$. This is the continuity axiom of preference.

Since this is an axiom, I don't think there is a proof.

Q1. I don't get the idea: how is this equivalent to $$(A \succeq B) \ \land \ (A \text{ sufficiently close to } C) \implies C \succeq B$$ ?

Similarly, what if only $$x^{\text{odd}} \succeq y$$ (and $$x^\text{even} \preceq y$$) and $$(x^{i})_{i \geq 1} \to x$$ holds, can't then $$x \succeq y$$?

Q2. In the above example, the axiom says it can be true although it's not necessary for $$x \succeq y$$ to hold true. In this regard, can I get an example to understand why the (original) axiom makes sense in the real world (or at all, in general) and this does not?

Edit: I am still not very clear with this; hence the answer below has not been accepted. I have added a few comments below the answer.

• Continuity is a preservation property. Suppose we have a mapping between two sets. Continuity ensures that a distance/magnitude/ordering in one set; is also sufficiently close in another set. Commented Jun 23, 2021 at 17:25
• @EB3112 I understand that; I am confused with the technicalities. I have written them below the answer, waiting for someone to explain that part to me. Commented Jun 23, 2021 at 18:34

Usually the continuity axiom has two parts

• if $$x_n \to x$$ and $$x_n \succeq y$$ for all $$n$$ then $$x \succeq y$$.
• if $$x_n \to x$$ and $$x_n \preceq y$$ for all $$n$$ then $$x \preceq y$$.

I don't get the idea: how is this equivalent to $$(A \succeq B) \wedge (A$$ sufficently close to $$C$$) $$\to$$ $$C \succeq B$$.

It is not. Take for example the case where preferences are strictly montonic which means that if $$x > y$$ then $$x \succ y$$ (more is always better). Then if we take two indifferent bundles $$x \sim y$$ we have that for any $$z < x$$ no matter how close to $$x$$ that $$z \prec y$$, as $$z \prec x \sim y$$.

What is true under continuity, however, is the following:

Th if $$x \succ y$$ then there exists an $$\epsilon > 0$$ such that for all $$z$$ that are $$\epsilon$$-close to $$x$$, we have $$z \succ y$$.

proof The proof is by contradiction. Assume not, then for all $$\varepsilon > 0$$ there is a $$z$$ which is $$\epsilon$$-close to $$x$$ and $$z \succeq y$$. Take a decreasing sequence of such $$\epsilon = 1, 1/2, 1/3,\ldots, 1/n, \ldots$$. Then for all $$n$$ there is a $$z_n$$ which is $$1/n$$-close to $$x$$ and $$z_n \succeq y$$. We have that $$z_n \to x$$, so by continuity $$x \succeq y$$, a contradiction.

Similarly, what if only $$x_{odd} \succeq y$$ (and $$x_{even} \preceq y$$) and $$(x_i)_{i \ge 1} \to x$$ holds, can't then $$x \succeq y$$?

Yes, this will indeed be the case.

To see this, assume that the condition hold. Take the subsequence of all odd terms $$x_1, x_3, x_5,\ldots$$ We have that for each $$x_i$$ in this term $$x_i \succeq y$$. Also, along this sequence $$x_i \to x$$, so by continuity $$x \succeq y$$.

The case is even stronger as we can also consider the subsequence of even terms $$x_2, x_4, \ldots$$. For each $$x_i$$ in this sequence $$x_i \preceq y$$ and $$x_i \to x$$. As such also $$x \preceq y$$. This means that both $$x \succeq y$$ and $$x \preceq y$$ giving that $$x$$ and $$y$$ are indifferent: $$x \sim y$$.

In the above example, the axiom says it can be true although it's not necessary for $$x \succeq y$$ to hold true. In this regard, can I get an example to understand why the (original) axiom makes sense in the real world (or at all, in general) and this does not?

I don't really understand this question.

Fix an $$y$$. Continuity then basically requires that there is not a sudden jump from the region where you are strictly better of than $$y$$, i.e. where $$x \succ y$$, to the region where you are strictly worse of than with $$y$$,i.e. where $$x \prec y$$.

In other words if you move in a continuous way from an element $$x$$ with $$x \succ y$$ to an element $$x'$$ with $$x' \prec y$$ then at some point you will find an element $$x''$$ on this path with $$x'' \sim y$$.

An alternative intuition is to look at the upper contour sets: $$U(y) = \{x: x \succeq y\}$$ and the lower contour sets $$L(y) = \{x: x \preceq y\}.$$ Continuity requires that both these sets are closed.

• Hi, thanks for your solution. Could you tell me why some articles define the continuity axiom as "A more preferable to to B and B sufficiently close to C implies A more preferable to C" and some articles state in the way you have. This is in regard to your answer to my first question. Commented Jun 23, 2021 at 15:07
• Wait, I reversed the statement there completely. It's only defined as in the previous comment. I thought it holds both ways. So it's ONLY $(A \succeq B) \land (B \to C) \implies A \succeq C$. Commented Jun 23, 2021 at 15:16
• Could you tell me if that is correct? Commented Jun 23, 2021 at 15:20
• @dictatemetokcus That's the second part of the definition of continuity. See the top of my answer.
– tdm
Commented Jun 23, 2021 at 15:21
• Do you mean $(A \succeq B) \land (A \to C) \implies C \succeq B$ is same as your first definition and $(A \succeq B) \land (B \to C) \implies A \succeq C$ as your second definition? Commented Jun 23, 2021 at 15:34