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Let $\hat{\mathbb{C}}$ denote the Riemann sphere. Let $f:B_1(0) \to \hat{\mathbb{C}}$ be continuous. If $f$ is continuous at $z$ and non-zero, then $1/f(z)$ is continuous at $z$ as well. My question is whether or not, when we define $1/f(z)=\infty$ for $f(z)=0$, we get continuity at $z$? What can be said for holomorphicity (that is, assume that $f$ is holomorphic and repeat my question).

My intuition for continuity is that $1/f(z)$ is not continuous. However, if $z_n \to z$, by the continuity of $f$, we get $\lim_{n \to \infty}\ f(z_n)=f(z)=0$, and so $\lim_{n \to \infty} 1/f(z_n)=\infty=1/f(z)$. This signals to me that $1/f(z)$ is therefore continuous. However, the $\epsilon$-$\delta$ argument will clearly fail, so $1/f(z)$ is not continuous? How are we supposed to treat continuity in this case? Similarly, what can be said for holomorphicity?

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    $\begingroup$ If $f$ is continuous at $z$ and $f(z) = 0$, then $1/f$ is continuous at $z$ as well if you define $1/f(z) = \infty$. That part is perfectly fine. $\endgroup$
    – David Gao
    Commented Feb 9 at 5:30
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    $\begingroup$ $1/f(z)$ is continuous and holomorphic at a zero of $f$ if you use the proper metric on $\hat {\Bbb C}$, the chordal metric: math.stackexchange.com/q/1159327/42969 $\endgroup$
    – Martin R
    Commented Feb 9 at 5:31
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    $\begingroup$ The reason $\epsilon-\delta$ fails because the Riemann sphere is not naturally a metric space. There are metrics on it that correctly generate the topology on $\hat{\mathbb{C}}$, as Martin already mentioned, but none of these metrics will match the natural metric on $\mathbb{C}$. In fact that is impossible since the Riemann sphere is compact, so any metric on it is necessarily bounded, whereas the metric on $\mathbb{C}$ is not. I would suggest to search for definitions of continuity on a topological space, which is the proper notion to use here. $\endgroup$
    – David Gao
    Commented Feb 9 at 5:34
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    $\begingroup$ Holomorphic is also fine, but you need to define what it means for a function with codomain in $\hat{\mathbb{C}}$ to be holomorphic. This can be done because the Riemann sphere is a complex manifold. In fact it is the projective line over $\mathbb{C}$. Once you make the correct definitions your question does have a positive answer. $\endgroup$
    – David Gao
    Commented Feb 9 at 5:38
  • $\begingroup$ What does "the $\epsilon$-$\delta$ argument will clearly fail" mean? $\endgroup$
    – Paul Frost
    Commented Feb 9 at 15:44

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