1. Dirichlet function proof

I'm having a hard time understanding this proof.

for each point $c$ of $R$.

each open interval of the form $(c - \frac{1}{n}, c+ \frac{1}{n})$, where $n \in N$
Contains ration $x_n$ and irrational $y_n$.

Ok so far, but then

Considering $\{x_n\}$ and $\{y_n\}$, we have $x_n \to c$ and $y_n\to c$, by the squeeze rule.

I don't understand what's going on here. As there are only two sequences, how are these two sequences being squeezed?

2. Re: Dirichlet function proof

Originally Posted by alyosha2
for each point $c$ of $R$.
each open interval of the form $(c - \frac{1}{n}, c+ \frac{1}{n})$, where $n \in N$
Contains ration $x_n$ and irrational $y_n$.
Ok so far, but then Considering $\{x_n\}$ and $\{y_n\}$, we have $x_n \to c$ and $y_n\to c$, by the squeeze rule. I don't understand what's going on here. As there are only two sequences, how are these two sequences being squeezed?
EACH sequence is being squeezed separately.

If $n\in\mathbb{Z}^+$ define ${I_n}(c) = \left( {c - {n^{ - 1}},c + {n^{ - 1}}} \right)$

Note that if $k\ge N$ then ${I_k}(c)\subseteq{I_N}(c)$. Thus $\left| {{x_k} - c} \right| < \frac{2}{k} \leqslant \frac{2}{N}$.

Do you see the squeezing now?

3. Re: Dirichlet function proof

I'm sorry, I'm still not sure I understand. We have a second sequence $x_k$ defined on a narrower interval. We are then saying this sequence is squeezed by the interval $x_n$on an interval of which the interval of $x_k$is a subset. But this only saying that if $x_n \to c$ then $x_k$ does, how does this tell us $x_n \to c$?

Also, why less $< \frac{2}{k}$ and not $< \frac{1}{k}$ as the sequence can never be more than that value from $c$?

4. Re: Dirichlet function proof

Originally Posted by alyosha2
I'm sorry, I'm still not sure I understand. We have a second sequence $x_k$ defined on a narrower interval. We are then saying this sequence is squeezed by the interval $x_n$on an interval of which the interval of $x_k$is a subset. But this only saying that if $x_n \to c$ then $x_k$ does, how does this tell us $x_n \to c$?
Also, why less $< \frac{2}{k}$ and not $< \frac{1}{k}$ as the sequence can never be more than that value from $c$?
I wrote it that way out of habit. Usually it is written as $\left| {{x_n} - {y_n}} \right| < \frac{2}{N},~~\forall n\ge N~.$

Now if $|x_n-c|<\frac{1}{N}$ then $|x_n-c|<\frac{2}{N}$.

If it is true that $(x_k)\to c$ then it is true that $(x_n)\to c$. The index makes no difference.

5. Re: Dirichlet function proof

I'm not sure I see. But here 's my understanding so far. Each rational/irrational number in each interval we define gets closer to the point on which we have defined the interval the narrower we make the interval. So a sequence on the larger interval will squeeze the sequence on the smaller interval. But I'm not sure how this helps. Are we saying something like we are creating a sequence of intervals and then extracting numbers from within those intervals to create a sequence of numbers that tend to the point around which the interval is defined?

6. Re: Dirichlet function proof

Originally Posted by alyosha2
Are we saying something like we are creating a sequence of intervals and then extracting numbers from within those intervals to create a sequence of numbers that tend to the point around which the interval is defined?
That is correct.