locally nonsatiated preferences

Question

what does this symbol mean in the discuss of locally nonsatiated preferences: $\varepsilon > 0$ and $||y-x||<\varepsilon$.

Answer 1

If you take the vector $\mathbf{x}$ as the center and an $\epsilon > 0$ sufficiently small as the radius, then the open ball (local neighborhood of the vector $\mathbf{x}$) is the set of all vectors $\mathbf{y} \in \mathbb{R}^{n}$ whose distance to the center $\mathbf{x}$ is less than $\epsilon$ measured w.r.t. the metric $\Arrowvert \cdot \Arrowvert $, i.e., more formally as

$$\mathbf{B}(\mathbf{x},\epsilon) := \big\{\mathbf{y} \in \mathbb{R}^{n}\,\Big\arrowvert\, \Arrowvert \mathbf{y} - \mathbf{x} \Arrowvert < \epsilon \big\}.$$

Notice that measuring a distance by a metric is only possible within a metric space or normed space like Hausdorff or Banach. Topological spaces do not have in general a metric, in this case the above concept must be generalized. Fortunately, all finite dimensional spaces have a canonical norm, i.e., the Euclidean norm.

Answer 2

$||v||$ is the norm symbol, it means the length of the vector $v$.

Answer 3

Depends on what exactly you mean with "this symbol":

1. $\varepsilon$ denotes a real number.
2. $>$ stands for the strictly-greater-than relation in the real numbers.
3. $0$ is the number zero, i.e., the neutral element with respect to addition.
4. $x$ and $y$ are consumption bundles, i.e., vectors of the same finite dimension with non-negative real entries.
5. $-$ is the minus sign, the inverse of the $+$ operation. The operation is performed element-wise for vectors.
6. $||v||$ is the norm (i.e., length) of the vector $v$. If $v\in\mathbb{R}^n$, then $||v||=(v_1^2+\cdots +v_n^2)^\frac{1}{2}$.
7. $<$ stands for the strictly-less-than relation in the real numbers.