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Let $f:\mathbb R^2\rightarrow \mathbb R$ be defined by $f(x,y)=1$ if $xy=0$ and $f(x,y)=2$ otherwise. Then find the set of points where $f$ is continuous.

I think $f$ is continuous on $\{(x,y):xy\neq 0\}$. Let $(x,y)\in \mathbb R^2$ be such that $xy\neq 0$. Then there is a neighbourhood of $(x,y)$ which does not intersect with x axis or y axis. So in that neighborhood the value of $f$ is 2 and hence $f$ is continuous at $(x,y)$. Am I correct?

Shreya Chauhan
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    You have proved that $f$ is continuous at $(x,y)$ if $x y\neq 0$. Now prove that $f$ is not continuous at $(x,y)$ if $x y= 0$. – Kavi Rama Murthy Nov 18 '20 at 08:50
  • @KaviRamaMurthy please answer this question or give hints, https://math.stackexchange.com/questions/3912353/show-that-max-x-in-a-b-fx-leq-fracb-a22-max-x-in-a-b?noredirect=1#comment8068843_3912353 – PNDas Nov 18 '20 at 08:52

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