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In reading this proof at PlanetMath:

Why is that induced topology the same as the metric topology? I only see that they are the base of some topology, not necessarily the metric topology.

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Let $\mathscr{T}$ be the metric topology, and let $\mathscr{U}$ be the topology generated by the base $\mathscr{B}$. That $\mathscr{U}\subseteq\mathscr{T}$ follows immediately from the fact that $\mathscr{B}\subseteq\mathscr{T}$: every member of $\mathscr{U}$ is a union of members of $\mathscr{T}$. It remains to show that $\mathscr{T}\subseteq\mathscr{U}$.

Suppose that $V\in\mathscr{T}$. For each $x\in V$ there is a positive integer $n_x$ such that $B(x,1/n_x)\subseteq V$; let $m_x=2n_x$. $\mathscr{B}_{m_x}$ covers $X$, so $x \in B(y+x,1/m_x)$ for some $B(y_x,1/m_x)\in\mathscr{B}_{m_x}$. For any $z\in B(y_x,1/m_x)$, $$d(x,z)\le d(x,y)+d(y,z)<2/m_x=1/n_x,$$ so $x\in B(y_x,1/m_x)\subseteq B(x,1/n_x)\subseteq V$. It follows that $$V = \bigcup_{x\in V} B\left(y_x,\frac1{m_x}\right) \in \mathscr{U},$$ since the sets $\displaystyle B\left(y_x,\frac1{m_x}\right)$ all belong to $\mathscr{B}$.

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Thanks for the explanation. – Steven Lin Sep 29 '11 at 1:39

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