proof verification: the set of irrational numbers is a dense subset of $\mathbb{R}$

Question

I am trying to prove the proposition that

The set of irrational numbers is a dense subset of $\mathbb{R}$

by using the Baire Category Theorem(referred as BCT from now on)

Let $\{E_n\}_{n=1}^{\infty}$ be a sequence of nowhere dense subsets of a complete metric space $X$. Then $\cup_{n=1}^{\infty} E_n$ has empty interior.

Let $\{r_n\}_{n=1}^{\infty}$ be a sequence of all rational numbers($\cup\{r_n\}=\mathbb{Q}$) in $\mathbb{R}$. Now consider $\{E_n\}_{n=1}^{\infty}$ where $E_n$ is a set of one rational number in $\mathbb{R}$. $E_n=\{r_n\}$ for each $n$. $E_n$ is nowhere dense since $\text{int}(\overline{E_n})=\phi$. $\cup_{n=1}^{\infty} E_n$ has empty interior by BCT.

This implies that for any rational $x$ and for any $r>0$, the open ball centered at $x$ with radius $r$, $B(x;r)$ contains an irrational number. It seems reasonable to conclude that the set of irrational numbers is a dense subset of $\mathbb{R}$.

I saw some more concise proofs exist such as

If $t$ is any irrational number and $r$ is a rational, then $\{r+\frac{t}{n}\}_{n=1}^{\infty}$ is a sequence of irrational numbers that converges to $r$.

Thank you.

Answer 1

For every real number $a$, just create the irrational sequence $\{a+\frac{e}{n}\} $, then you got your proof simply.

Answer 2

I'll show you an easier way way to prove that result.

A set $B$ is dense in $(X,\tau)$, if every non empty open interval intersects $B$ non-trivially, this is: $\forall U \in \tau, U \cap B \neq \emptyset$.

Let $\mathbb I$ be the set of all irrational numbers. Let's use $\cal B = \{(a,b) \subset \mathbb R\}$ as a basis for $\mathbb R$.

Let $U \in \tau$. Then $U = \bigcup_{j \in J} (a_j,b_j)$, for some index set $J$. Let $i \in J$. We have that between 2 real numbers there is allays an irrational number (You should prove this if you didn't already). So $\exists \alpha \in \mathbb I: \alpha \in (a_i, b_i)$. So $\alpha \in \bigcup_{j \in J} (a_j,b_j) = U$. Thus $\forall U \in \tau, U \cap \mathbb I \neq \emptyset$ concluding that $\mathbb I$ is dense in $\mathbb R$.