Understanding the Continuity axiom of preference

Question

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?

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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.

Answer 1

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.