What’s special about Thomae’s function? That one here:[math: t(x)={1 if x=0;1n if x is a rational number, x=m/n in lowest terms, n0;0 if x is irrational.]t(x)= \begin{cases} 1 \quad & ext{ if $x=0$;} \ \frac{1}{n} \quad & ext{ if $x$ is a rational number, $x=m/n$ in lowest terms, $n0$;} \ 0 \quad & ext{ if $x$ is irrational.} \end{cases}

Question

What’s special about Thomae’s function? That one here:[math: t(x)={1 if x=0;1n if x is a rational number, x=m/n in lowest terms, n>0;0 if x is irrational.]t(x)= \begin{cases} 1 \quad &	ext{ if $x=0$;} \ \frac{1}{n} \quad &	ext{ if $x$ is a rational number, $x=m/n$ in lowest terms, $n>0$;} \ 0 \quad & 	ext{ if $x$ is irrational.} \end{cases}

BlackTom AI · Accepted Answer

The question asks about the special properties of Thomae’s function t(x), defined as t(0)=1, t(x)=1/n if x is rational x=m/n in lowest terms (n>0), and t(x)=0 if x is irrational. This is a classic example in real analysis showing a function that is continuous at all irrationals but discontinuous at all rationals.

Option a: It is continuous at every irrational x but discontinuous at every rational x. This statement matches the standard result: near any irrational, you can find rationals with arbitrarily large denominators, whose values 1/n tend to 0, and since t is 0 at irrationals, the limit is 0, matching t(x). At any rational p/q, the function value is 1/q, but one can approach p/q by a sequence of irrationals with t-values 0, so the limit is 0 which differs from t(p/q)=1/q, hence discontinuity at rationals. This option is correct.

Option b: It is continuous everywhere. This is false because at every rational x=p/q we have t(p/q)=1/q but nearby irrationals have t-values 0, so the limit cannot equal 1/q. Thus not continuous at rationals.

Option c: It is continuous at every rational x but discontinuous at every irrational x. This is the opposite of the truth. Thomae’s function is not continuous at rationals, and it is continuous at irrationals, so this statement misstates the continuity behavior.

Option d: It is discontinuous everywhere. This is incorrect because we just noted that the function is continuous at every irrational x, so it is not discontinuous everywhere.

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