I am trying to show that Thomae's function is continuous at every irrational point (I have already shown discontinuous at every rational point), and for that – following this answer Prove continuity/discontinuity of the Popcorn Function (Thomae's Function). – I would like to show that for an irrational number $x$, given $N\in{\mathbb{N}}$, there exists $\epsilon>0$ so that all rationals $\in{V_\epsilon{(x)}}$ have denominator larger than N.
Proof Irrational number $x$, given $N\in{\mathbb{N}}$ exists $\epsilon>0$ so all rationals $\in{V_\epsilon{(x)}}$ have denominator larger than N.
continuitygeneral-topologyirrational-numbersrational numbersreal-analysis
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Detailed sketch of the proof:
Since $\mathbb{Q}\cap[0,1]$ is countable, we can write it $\mathbb{Q}\cap[0,1]=(x_n)_{n\in\mathbb{N}}$, where each $x_n$ has a canonical form $x_n=\frac{p_n}{q_n}$, $p_n\wedge q_n=1$. We then define $f_n$ by considering only the $n+1$ first terms of the sequence $(x_k)$, and putting the "right" value on them (i.e, $f_n(x_k)=\frac{1}{q_n}$, $0\leq k\leq n$).
To make $f_n$ continuous, we then make it "affine by parts": each $x_k$ is surrounded by two elements of $\{0,x_1,\dots,x_n,1\}$ (finite number of elements), we can pick the max $\underline{x}_k$ of those strictly less, the min $\bar{x}_k$ of those strictly greater, and then go from 0 to $\frac{1}{q_n}$ on the interval $\left[x_k-\frac{x_k-\underline{x}_k}{2^n}, x_k\right]$, then from $\frac{1}{q_n}$ to 0 on the interval $\left[x_k, x_k+\frac{\bar{x}_k-x_k}{2^n}\right]$ (cf. the very bad picture below).
Since each rational $q$ is one of the $x_n$ and thus will, for $k\geq n$, be be "considered" by $f_k$ and its image put to the desired value, $f_k(q)\to f(q)$ for all rationals in $[0,1]$. The problem is for the irrationals. Pick $r\in\mathbb{R}\setminus\mathbb{Q}$ arbitrary; in order to prove that $f_k(r)\to0$, we can, by contradiction, suppose it doesn't.\ It means that there is an $\varepsilon>0$ such that, for an infinity of $k$, $f_k(r)>\varepsilon$. Pick the subsequence of such $k$: it defines a sequence $(f_{\varphi(k)}(r))_k$ of values always strictly greater that $\varepsilon$.
But that means that there is a corresponding sequence of rationals always very close to $r$ (one for each $f_{\varphi(k)}$, otherwise $f_{\varphi(k)}(r)=0$). By very close, it actually means (closer than $\frac{\bar{q}-q}{2^n}\leq 2^{-\varphi(k)}$ for some rational $x$. In other terms, we have a sequence of rationals converging (since $2^{-\varphi(k)}\to0$, because $\varphi(k)\to\infty$ (it is a subsequence, so $\varphi(k)\geq k$ for every $k$).
Then, we can use the fact that if a sequence of rationals $\frac{p_n}{q_n}\to r$ irrational, necessarily $p_n\to \infty$, $q_n\to \infty$ (it is rather simple to prove*). And thus in particular $\frac{1}{q_n}\to 0$. Since $f_{\varphi(k)}(r)$ is always less than the "peak", which is the $\frac{1}{q}$ of the "close rational associated with $f_{\varphi(k)}$" mentioned earlier, $f_{\varphi(k)}(r)\to0$ when $k\to\infty$. Which is a contradiction, because it is supposed not to go under $\varepsilon>0$.
Hence, $f_k(r)\to 0$, and $f_n$ converges pointwise to $f$.
${}$* for example, by contradiction, supposing some subsequence of $p_n$ (or $q_n$) is bounded, applying Bolzano—Weierstrass' theorem to it to have a converging subsequence, and then considering the corresponding subsequence of $q_n$ which must be itself, then, bounded because of the convergence to $r$, and again apply Bolzano—Weiestrass, and find finally a sub-sub-sequence of $p_n$ and $q_n$ such that both converge to some integers $p$ and $q$, which in turn implies that $r=\frac{p}{q}\in\mathbb{Q}$, which can be seen as a problem for an irrational.
Here's a proof that I accept is cheating at certain level, but it works for sure-
We know that the Thomae's function is discontinuous at rationals and continuous at irrationals. So, the negative of this function should also satisfy these same properties. So, $1$ added to the negative of this function should also satisfy these.
Since the given function is $f(x)=1-T(x)$ where $T(x)$ is the Thomae's function, $f(x)$ must be discontinuous at rationals and continuous at irrationals.
Best Answer
Since there are only finitely many rational points in $[0,1]$ whose denominator is not larger than $N$ there has to be a $\varepsilon$ so small that $(x-\varepsilon,x+\varepsilon)$ contains none of them; just take$$\varepsilon=\min\{|x-y|\,:\,y\text{ is rational and its denominator is}\leqslant N\}.$$