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A pseudometric (aka. pseudo-distance) is a metric except that maybe $x \neq y$ but $d(x,y) = 0$.

Consider a family $(d_a)_{a \in A}$ of pseudometrics on a set $E$. For each $x \in E$ and each finite family $(a_j)_{j=1\dots m} \subset A$ and finite family $(r_j)_{j = 1\dots m} \subset \Bbb{R}_{\gt 0}$, define the "ball" $B(x; (a_j), (r_j)) = \{y\in E : d_{a_j}(x,y) \lt r_j \text{ for } 1 \leq j \leq m\}$.

Let $\mathfrak{D}$ denote the set of all subsets $U$ of $E$ such that for all $x \in U$ there's one of those "balls" $B(x; \dots) \subset U$. Then it's immediately verified that $\mathfrak{D}$ is a topology on $E$. I see the $\varnothing, E \in \mathfrak{D}$ part, but having trouble with the finite intersection of two $U$'s: $U_1 \cap U_2 \in \mathfrak{D}$ when both $U_1, U_2$ are. How do I choose the finite families $(a_j), (r_j)$ so that the resulting ball $B(x; \dots) \subset U_1 \cap U_2$, for any $x \in U_1 \cap U_2$?

Clearly we can't intersect the family for $x \in U_1$ with the family for $x \in U_2$ can we? As the intersection could be empty and that doesn't make much of a ball. This problems got me by the balls. lol

Thanks.

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Each $B(x; (a_j), (r_j)) = B(x; a_1, r_1) \cap B(x; a_2, r_2) \cap \dots \cap B(x; a_m, r_m)$ clearly. Take some balls $B_1 \subset U_1$ and $B_2 \subset U_2$ each containing $x$ and intersect them to get another such ball. So it was the union of the families $(a_j, r_j)^1,(a_j, r_j)^2$ after all!

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