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Can a 2-dimensional scalar field have a discontinuous contour curve? How about contour curves that intersect -- possible?

On a related note: can a vector field have a domain that is not defined over a continuous region??

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The title doesn't match the actual question. – Hans Lundmark Sep 30 '12 at 8:52
Even smooth scalar fields can have contour curves that intersect or consist of several disconnected pieces. Consider for example $xy=0$ or $xy=1$. – Hans Lundmark Sep 30 '12 at 9:01
The concept of scalar field I'm being given is a physical one: "An n-dimensional scalar field is a distribution of scalar values in n-dimensional space, and is represented mathematically by a scalar function of n variables." But you guys seem to treat "scalar field" as being synonymous with "scalar function". – Ryan Sep 30 '12 at 9:04
@Ryan: Is "made up of points that don't join up" an euphemism for "discontinuous"? – Rod Carvalho Sep 30 '12 at 9:06
@rod Oops, yes; I have duely edited my question for concision. – Ryan Sep 30 '12 at 9:13
up vote 2 down vote accepted

Yes. For example, the scalar field $f (x,y) := x y$ has discontinuous contour curves. Note that $f (x,y) = 1$ yields two hyperbolas:

Two hyperbolas

Animated plot courtesy of Wikipedia.

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Thank you, Rod. – Ryan Sep 30 '12 at 9:24

Yes, there are many ugly functions.. it also depends what do you mean by a 2 dimensional scalar field (because one usually consider them at least continuous by definition).

So, for example, take the following function: $$(x,y) \mapsto \left\{ \begin{matrix} 0 & \text{if }x,y\in\mathbb Q \\ 1 & \text{else} \end{matrix} \right. $$

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"it also depends what do you mean by a 2 dimensional scalar field (because one usually consider them at least continuous by definition)." This is precisely what I'm trying to figure out: are scalar fields continuous functions by definition?? My intuition tells me yes (and also, the concept of contour curves, etc., all seem to point towards this), yet I can't find any definition that mandates that scalar fields be continuous functions. – Ryan Sep 30 '12 at 8:57

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