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With respect to the usual metric of C

this $${z \in C: 1< |z| <= 2}$$ is not closed, however with another metric such as

$$d(z,w)$$ where $$d(z,w) = 0,z=w$$ and $$d(z,w) = |z| +|w|, z\neq w$$ this subset is closed? Can someone please explain why?

Saurav Sathnarayan
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    Please use MathJax instead of images. https://math.stackexchange.com/help/notation – Haris Mar 06 '24 at 20:18
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    Well yes, whether a subset is closed or not is not inherent in the set "itself" "as such", but depends on what metric (or more generally, topology) one puts on the ambient space. So to be precise, one should always say "closed with respect to this and that metric". Congratulations for understanding that. (Often there is only one relevant / interesting metric, and it is understood from context, and people would not say it each time.) – Torsten Schoeneberg Mar 06 '24 at 20:19
  • @torstenSchoeneberg Thank you for that. However it does not answer my question. – Saurav Sathnarayan Mar 06 '24 at 20:26
  • So your question is only why that specific set is closed w.r.t. that specific metric $d$? Then I was misled by your question title. – Torsten Schoeneberg Mar 06 '24 at 20:46
  • Notice in the latter case distances have a minimum. – TurlocTheRed Mar 06 '24 at 21:31
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    Answer: why not? Note for instance that for the discrete metric, every subset is closed. So, what is your question, actually? Does asking "why" mean "help me to prove that"? If so, please show your own attempts first, and change your title to a less misleading one. – Anne Bauval Mar 06 '24 at 21:57

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To answer why that specific set is closed in that specific metric, hint: consider a $z \neq 0$ and let $0 <r <|z|$. What is the open ball of radius $r$ around $z$? From this you should conclude that many sets are open w.r.t. this metric, including the complement of yours.

Torsten Schoeneberg
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