Domain of f(x,y): Open or Closed?

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SUMMARY

The domain of the function \(f(x,y)=\frac{\sqrt{ln(x^{2}+y^{2}+1)}}{\left | x \right |+\left | y \right |+\sqrt[4]{xy-1}}\) is classified as a closed and unbounded set. The confusion arises from the distinction between closed sets and boundedness; a set can be closed even if it extends to infinity. The discussion clarifies that the definition of a closed set does not require it to be bounded or that every point is an interior point, as exemplified by the entire plane being a closed set.

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Yankel
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Hello all,

I am trying to determine if the domain of the function:

\[f(x,y)=\frac{\sqrt{ln(x^{2}+y^{2}+1)}}{\left | x \right |+\left | y \right |+\sqrt[4]{xy-1}}\]

Is an open set or closed set and if it's bounded.

The domain is in the attached graph.

View attachment 2468

The book say it is closed and unbounded. I wonder, how can it be closed, when it goes to infinity ?

I may be confusing boundary with open/close, but shouldn't it be open if it goes to infinity, or is it enough to say that since every point is interior it is closed ?

thanks !

Edit: What I mean is, isn't it like sets of 1 variables, were we always write [a,infinity) since infinity can't be closed ?
 

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You need to review the definition of "closed". It does not mean that the set is bounded or that every point is an interior point. For example, the whole plane is closed.

Yankel said:
What I mean is, isn't it like sets of 1 variables, were we always write [a,infinity) since infinity can't be closed ?
Interesting remark. A parenthesis as opposed to a square bracket means that the boundary does not belong to the interval. Infinity is not a real point, so by convention it is considered to not belong to any interval. And yes, for a finite interval both boundaries belong to it iff it is closed. However, this does not hold for infinite intervals.
 

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