In my textbook, the Bolzano-Weierstrass Theorem is stated as:

Bolzano-Weierstrass Theorem: A bounded sequence $\left(x_{n}\right)\in\mathbb{R^{n}}$ has a convergent subsequence.

Does this hold true for metric spaces $\left(\mathbb{R^{n}},d\right)$, where $d$ is an arbitrary metric? Or does $d$ have to be the standard Euclidean metric?


No, it does not hold for any metric. Take for example $d(x,y)=|\arctan(x)-\arctan(y)|$, then any sequence in $\mathbb{R}$ is bounded with respect to $d$.

However, Bolzano-Weierstrass Theorem holds if the metric is induced by a norm because in $\mathbb{R}^n$ every norm is equivalent to the euclidean one.

  • $\begingroup$ Or we can take the bounded metric $\overline{d}(x, y) = \min(d(x, y), 1)$ where $d$ is the Euclidean metric. $\endgroup$ – Alex Vong May 23 '17 at 7:52

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