Can we draw this function? $f\colon\mathbb{R}\to\mathbb{R}$, given by $$f(x) = \left\{\begin{array}{ll} 1 &\mbox{if $x\in\mathbb{Q}$;}\\ 0 &\mbox{if $x\notin\mathbb{Q}$.} \end{array}\right.$$
Can we draw the characteristic function of the rationals?
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geometry
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2Depends what you mean by "draw"; Plank's constant would get in the way if you try to accurately picture it; the rationals and the irrationals are both dense on the line, so any drawing in which a point in the graph has any area (no matter how small) would simply "look" like two horizontal lines, one at height $0$ and one at height $1$. – 2011-04-15
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1@Arturo @Ross Millikan: Actually, I find the idea of defining what it means for a function to be "drawable" to be kind of interesting. I'd submit that a good definition would be that you could low-pass filter the function and write the result as a piecewise smooth function. – 2011-04-15
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0If you want to draw dotted where and how? so your observation doesnt give any further information, just discontuity of function. and an interesting drawing-curve must give more information than that. – 2011-04-15
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0I know that it's discontinuous – 2011-04-15
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0@Mopzer: Is your first comment addressed at Ross's answer? If so you should comment on his answer, not your question. If it's addressed at me, I don't understand the comment. – 2011-04-16
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0@Mopzer: There *isn't* that much information asbout this function: it takes exactly two values, both of them on dense sets; it is discontinuous everywhere, but that is not something draw a "drawing" will let you see. There are **physical limits** to what you can represent pictorially, and an accurate representation of this function is well beyond those limits. – 2011-04-16
1 Answers
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It depends upon what you mean by "draw". You could draw a dotted line along $y=0$ and another along $y=1$ with the dots meaning that the function is not continuous and so some of the points you (seem to) have covered are not part of the function. If by draw you mean a continuous curve in the plane, no you cannot, because the function is not continuous.
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5I would draw the line at 0 a bit thicker because it contains so much more points... ;-) – 2011-04-15
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0@Fabian: for a human that's might be right, all that depends on limit of our eyes and density of both rationals and irrationals in real set. – 2011-04-18
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0@Arturo: Sorry. I'm a new user, I didnt know how to specify interested person, I didnt have any idea indeed. I asked Ross – 2011-04-18
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0@Fabian I was thinking about it, maybe the line at $1$ should be invisible (not even be drawn)? – 2018-07-19
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0@Ovi: the point of drawing something is to give an idea of what it is. It is important that the value of the function is $1$ for some points. We don't want to give the impression the function is $y=0$. It is true the rationals are measure zero in the reals, but I think the two dotted lines is the best representation. Fabian's idea to have the one at $y=0$ heavier seems reasonable to me but not necessary. – 2018-07-19
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0Ok makes sense ${}$ – 2018-07-19