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For each $i\in I$, $$d_i:X^2\to \Bbb R$$ is a metric and $\mathcal T$ is the coarsest topology on $X$ containing all topologies produced by the $d_i$ .

For $U\subseteq X$ we have: $$\forall x \in U,~\exists i\in I,~\exists r>0,~~B_x^{d_i}(r)\subseteq U$$ where $B_x^{d_i}(r)$ is the open ball with center $x$ and radius $r$.

Is $U$ open in $(X,\mathcal T)$?


Edit:

Is $U$ open in some $(X,d_i)$?

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2 Answers 2

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Yes. For any $x\in X,i\in I$ and $r>0$, $B_x^{d_i}(r)\in\mathcal{T}_i$, the topology generated by $d_i$, and $\mathcal{T}_i\subseteq\mathcal{T}$, so $B_x^{d_i}(r)\in\mathcal{T}$ for each $x\in X,i\in I$ and $r>0$. In particular, each point of $U$ has a $\mathcal{T}$-open nbhd contained in $U$, so $U\in\mathcal{T}$.

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  • $\begingroup$ then, is $U$ open in some $(X,d_i)$? $\endgroup$
    – user59671
    Mar 1, 2013 at 13:14
  • $\begingroup$ @CutieKrait: Not necessarily, because you might need a different $i$ for different points of $U$. If you can use the same $i$ for each $x\in U$, then yes, $U$ is open in that $\mathcal{T}_i$. $\endgroup$ Mar 1, 2013 at 13:16
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For all $x \in U$, there exist $i \in I$ and $r_x>0$ such that $B_x^{d_i}(r_x) \subset U$, so $U= \bigcup\limits_{x \in U} B_x^{d_i}(r_x)$. But each $B_x^{d_i}(r_x)$ is open for the metric $d_i$, and a fortiori for $\mathcal{T}$ by definition.

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