Prove that this $\epsilon$-ball intersects $C$

Question

Let $C=\{(x,\ y)\in\mathbb{R}^2\mid x,\ y\in\mathbb{Q}\text{, and }|x|+|y|\le 1\}$

Prove that $C$ is not closed in $\mathbb{R}^2$.

My attempt:

To do this, I plan to show that $\displaystyle\left(\frac{1}{\pi},\ \frac{1}{\pi}\right)\in\overline{C}$.

Since $\displaystyle\left(\frac{1}{\pi},\ \frac{1}{\pi}\right)\notin C$, that would prove that $\overline{C}\ne C$.

Let $\displaystyle B((x_0,\ y_0),\ \epsilon)$ be an $\epsilon$-ball containing $\displaystyle\left(\frac{1}{\pi},\ \frac{1}{\pi}\right)$. How do I prove that this $\epsilon$-ball intersects $C$?

Answer 1

I would suggest in this way: Because $(1/\pi,1/\pi)\in\{|x|+|y|<1\}$, and there exists a sequence $p_{n}\in{\bf{Q}}$ such that $p_{n}\rightarrow 1/\pi$, so we have eventually that $(p_{n},p_{n})\in C$, but the limit point $(1/\pi,1/\pi)\notin C$, so $C$ is not closed.

Answer 2

The answer may depend on how much you know (or accept) the facts.

If you accept the following two facts, then the answer is easy.

Fact 1. The decimal expansion of an irrational number never repeats or terminates. Say $$ \begin{align*} \frac{1}{\pi} &= 0.31830988618379067153776752674502872406891929148091\dotsb \\ &= 0.a_1a_2a_3a_4a_5\dotsb = \sum_{n=1}^\infty\frac{a_n}{10^n} \end{align*} $$

Fact 2. (Archimedean property) For any real number $\epsilon>0$, there exists a natural number $N$ such that $\frac{1}{\epsilon}<N$.

Given a real number $\epsilon>0$, it suffices to find a rational number $q$ such that $|\frac{1}{\pi}-q|<\frac{\epsilon}{\sqrt{2}}$. Then $$ |(\tfrac{1}{\pi},\tfrac{1}{\pi})-(q,q)|=\sqrt{2(\tfrac{1}{\pi}-q)^2}=\sqrt{2}|\tfrac{1}{\pi}-q|<\epsilon $$ so that $(q,q)$ is included in the $\epsilon$-ball centered at $(\tfrac{1}{\pi},\tfrac{1}{\pi})$.

Now let us find a rational number $q$ such that $|\frac{1}{\pi}-q|<\frac{\epsilon}{\sqrt{2}}$. By Fact 2, there exists a natural number $N$ such that $\frac{1}{\epsilon/\sqrt{2}}<N$.

By Fact 1, we can write the decimal expansion of $\frac{1}{\pi}=0.a_1a_2a_3a_4a_5\dotsb$. Set $$ q = 0.a_1a_2a_3\dotsb a_N = \sum_{n=1}^N\frac{a_n}{10^n} $$ Then we have $$ |\tfrac{1}{\pi}-q|=\sum_{n=N+1}^\infty\frac{a_n}{10^n}<\frac{1}{10^N}<\frac{1}{N}<\frac{\epsilon}{\sqrt{2}} $$

Answer 3

By the definition of the reals we have

(i). $\Bbb R$ is order-dense in itself. That is, if $a,b\in \Bbb R$ with $a<b,$ then the real interval $(a,b)$ is not empty.

(ii). $\Bbb Q$ is dense in $\Bbb R.$ That is, $J\cap \Bbb Q\ne \emptyset$ for every non-empty open real interval $J.$

So for every $n\in \Bbb N$ let $x_n$ be a rational member of the real interval $(\frac {1}{\pi}(1-2^{-n}),\frac {1}{\pi}).$ The ordered pair $p_n=(x_n,x_n)$ belongs to $C,$ and the sequence $(p_n)_{n\in \Bbb N}$ converges to $(\frac {1}{\pi},\frac {1}{\pi}),$ which is not in $C.$