Can
$\mathbb{R}^2 \simeq \mathbb{C}$ and because $\mathbb{R}^2$ is complete $\implies \mathbb{C} $ is complete
be an argument to show thaf $\mathbb{C}$ is complete? Or can you give me a proof please?
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MathLearner
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1It depends what you mean by $\simeq.$ Hint: completeness is relative to a metric. – Anne Bauval Mar 21 '23 at 22:04
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I say yes. But you should show that any Cauchy sequence $z_n \subset \mathbb{C}$ has a limit in $\mathbb{C}.$ Hint: real and imaginary parts. – Sean Roberson Mar 21 '23 at 22:05
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You have to show they're isomorphic as metric spaces for that argument to work. – eyeballfrog Mar 21 '23 at 22:06
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Yes eyeballfrog, this is what I meant. – Anne Bauval Mar 21 '23 at 22:10
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@eyeballfrog, what mean that they're isomorphic at metric space? – MathLearner Mar 21 '23 at 22:18
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1That you preserve the metric/norm. So it would actually here suffice that they are isomorphic as normed spaces, and so showing that, if we call your iso $f$, that $x$ has the same norm as $f\left(x\right)$. – Felix Mar 22 '23 at 07:34