The continuity of Thomae's Function

The Thomae's function is an area of great interest in real analysis. It is given by:

In this note, I will discuss the continuity of the function at different points in the interval [0,1]. First we shall check whether this function is continuous at rational points in the domain or not. Let cQc\in Q, the set of rationals. Then f(c)0f(c)\neq 0. Now we attempt to choose a sequence (xn)n1{ \left( { x }_{ n } \right) }_{ n\ge 1 } of irrationals in [0,1] which converge to c. Clearly, such a sequence can always be chosen owing to the density property of irrationals. Then (f(xn))n1{ \left( { f(x }_{ n }) \right) }_{ n\ge 1 } is a constant sequence (the zero sequence), which is clearly not convergent to f(c). So the function is discontinuous at every rational point. Now, let us choose an irrational point, say, b. So, f(b)=0f(b)=0 Examining the continuity of the function at irrational points is a bit tricky and needs use of the εδ\varepsilon -\delta criterion of continuity. We see that the 1q\frac { 1 }{ q } thing implies the need of using the Archimedean property in proving (or disproving) the continuity. According the εδ\varepsilon -\delta definition, for an arbitrary ε>0\varepsilon >0, we have to find a δ>0\delta >0 such that: xb<δf(x)f(b)=f(x)<ε\left| x-b \right| <\delta \\ \Longrightarrow \left| f(x)-f(b) \right| =\left| f(x) \right| <\varepsilon Now, if we choose such an x from the irrationals then f(x)=0f(x)=0, and the above condition holds for any δ>0\delta >0. However the problem comes if we are to choose such an x from the rationals. But aha, the Archimedean Property comes to rescue. We see, that for any ε>0\varepsilon >0 n0N,s.t.nn0,nN,1n<ε\exists { n }_{ 0 }\in N,\quad s.t.\quad \forall n\ge { n }_{ 0 },\quad n\in N,\frac { 1 }{ n } <\varepsilon . Thus we see that each of 11,12,13,..,1n01\frac { 1 }{ 1 } ,\frac { 1 }{ 2 } ,\frac { 1 }{ 3 } ,..,\frac { 1 }{ { n }_{ 0 }-1 } is greater than ε\varepsilon . Again, we see, that for any nNn\in N there are (n+1)(n+1) rationals lying in [0,1], which are mapped to 1n\frac { 1 }{ n } under the Thomae's function, viz., 0n,1n,2n,3n,..,nn\frac { 0 }{ n } ,\frac { 1 }{ n } ,\frac { 2 }{ n } ,\frac { 3 }{ n } ,..,\frac { n }{ n } . Thus for the n0{ n }_{ 0 } that we have found, we see that there are (n1)(n-1) rationals each with numerator 1, that are greater than or equal to ε\varepsilon , and hence there are (1+1)+(2+1)+(3+1)+..(n01+1){(1+1)+(2+1)+(3+1)+..({ n }_{ 0 }-1+1)} many rationals which disturb the inequality xb<δ\left| x-b \right| <\delta and affect our choice of δ\delta . Thus we are to choose δ\delta so as to exclude all these disturbing rationals from the open interval (bδ,b+δ)\left( b-\delta ,b+\delta \right) . This can be effected by choosing:δ=12min{mnb,wherem=0(1)n,n=1(1)(n01)}\delta =\frac { 1 }{ 2 } min\left\{ \left| \frac { m }{ n } -b \right| ,\quad where\quad m=0(1)n,n=1(1)({ n }_{ 0 }-1) \right\} ,i.e., making the gap of δ\delta and b even lesser than the gap between b and the disturbing rational nearest to b. Thus we see, that for an arbitrary ε>0\varepsilon >0, we can find a δ>0\delta >0 such that: xb<δf(x)f(b)=f(x)<ε\left| x-b \right| <\delta \\ \Longrightarrow \left| f(x)-f(b) \right| =\left| f(x) \right| <\varepsilon . Thus Thomae's function is continuous at irrationals.

Note by Kuldeep Guha Mazumder
4 years, 3 months ago

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Is the function differentiable at the irrationals?

What if we change the definition to f(pq)=1qn f(\frac{p}{q} ) = \frac{1} { q^n} ? What can we say for different values of nn?

Calvin Lin Staff - 4 years, 3 months ago

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Oh! Yes yes you gave me a nice point to write another note on..Thomae's function is not differentiable at any real point. I will supply a proof of that shortly.

Kuldeep Guha Mazumder - 4 years, 3 months ago

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