0

Consider the sequence $(x_n)_{n\in\mathbb{N}} = 2+ \frac{1}{n}$ in $\mathbb{C}$ with the metric $$d(x,y) = \begin{cases} |x-1| + |y-1|&\text{if $x\neq y$}\\ 0 &\text{if $x=y$} \end{cases} $$ Then one is able to show that this is a bounded sequence but it cannot possess a convergent subsequence. So my question is: why is this not a contradiction to the Bolzano Weierstraß Theorem for $\mathbb{R}^n$ with $\mathbb{C}$ understood as $\mathbb{R}^2$?

Robert Z
  • 145,942
HelloEveryone
  • 449
  • 1
    $d$ is not a metric because $d(x,x)$ should be zero. – Martin Argerami May 11 '23 at 17:47
  • 1
    Even if $d$ were a metric, the Bolzano−Weierstraß theorem applies for the usual topology of $\Bbb R^n$. – nejimban May 11 '23 at 17:48
  • 1
    The metric that you propose is not equivalent to the topology of $\mathbb R^n$. – Kroki May 11 '23 at 17:50
  • Choose $A=[\pi+e, \pi e]$ in $(\Bbb{Q}, d_{\text{std}}) $.Then $A$ is closed and bounded but not compact or choose any infinite discrete space. – Sourav Ghosh May 11 '23 at 17:58
  • 1
    The subspace $\mathbb{C} \setminus {1}$ of $(\mathbb{C}, d)$ is discrete. It's not surprising that a bounded sequence in a discrete space doesn't have a convergent subsequence. (As others have said, Bolzano-Weierstrass doesn't apply to every metrizable topology on $\mathbb{R}^n$, just the usual one) – Brian Moehring May 11 '23 at 18:00
  • Simple example, choose $X=\Bbb{R}$ and $d(x, y) =0 $ if $x=y$ and $1$ otherwise. $\$ In this metric $[0, 1]$ is closed and bounded but not compact. – Sourav Ghosh May 11 '23 at 18:26
  • Thanks for the clarification! – HelloEveryone May 11 '23 at 18:30

0 Answers0