Given are the following two statements ($x_r$ denotes a sequence of rational numbers, and $x_i$ denotes a sequence of irrational numbers): $$1. \forall x_r \to 1 \implies f(x_r) \to f(1)$$ $$2. \forall x_i \to 1 \implies f(x_i) \to f(1)$$

Are these sufficient to conclude that $$\forall x_n \to 1 \implies f(x_n) \to f(1) $$ where $x_n$ denotes sequence of real numbers.

If one needs the $f$, here it is:$$f(x) = \begin{cases} x-1 & x \in \mathbb{Q} \\ 0 & x \in \mathbb{Q}^{c} \end{cases}$$


3 Answers 3


Yes, it is.

Hint: Given the sequence $x_n$, suppose there are an infinite number of both rational and irrational terms (else the proof is trivial). Define two subsequences by taking the rational terms in the first one and the irrational terms in the second one, then do the usual $\epsilon$-$\delta$ estimations to prove that the whole sequence indeed converges to $1$.


You are expressing yourself very poorly, but the idea is correct. Indeed, if $f\colon\mathbb{R}\longrightarrow\mathbb R$ is such that

  • for every sequence $(x_n)_{n\in\mathbb N}$ of rational numbers, $\lim_{n\to\infty}x_n=1\implies\lim_{n\to\infty}f(x_n)=1$;
  • for every sequence $(x_n)_{n\in\mathbb N}$ of irrational numbers, $\lim_{n\to\infty}x_n=1\implies\lim_{n\to\infty}f(x_n)=1$

then it is true that, for every sequence $(x_n)_{n\in\mathbb N}$ of real numbers,$$\lim_{n\to\infty}x_n=1\implies\lim_{n\to\infty}f(x_n)=1.$$

In fact, given such a sequence, there are three possibilites:

  1. $x_n\in\mathbb Q$ is $n$ is big enough. Then, for some $N\in\mathbb N$, $n\geqslant N\implies x_n\in\mathbb Q$. Since the limit of the sequence $\bigl(f(x_n)\bigr)_{n\geqslant N}$ is $1$, $\lim_{n\to\infty}f(x_n)=1$.
  2. $x_n\in\mathbb{R}\setminus\mathbb Q$ is $n$ is big enough. Then, for some $N\in\mathbb N$, $n\geqslant N\implies x_n\in\mathbb{R}\setminus\mathbb Q$. Since the limit of the sequence $\bigl(f(x_n)\bigr)_{n\geqslant N}$ is $1$, $\lim_{n\to\infty}f(x_n)=1$.
  3. Otherwise, let $N_1=\{n\in\mathbb{N}\,|\,x_n\in\mathbb{Q}\}$ and let $N_2=\{n\in\mathbb{N}\,|\,x_n\in\mathbb{R}\setminus\mathbb{Q}\}$. Then $\lim_{n\in N_1}x_n=\lim_{n\in N_2}x_n=1$. Therefore, $\lim_{n\in N_1}f(x_n)=\lim_{n\in N_2}f(x_n)=1$ and, since $\mathbb{N}=N_1\cup N_2$, $\lim_{n\in\mathbb N}f(x_n)=1$.

Yes, this is true.

Let $(x_n)_{n \in \mathbb{N}}$ a sequence with $\lim_{n \to\infty}=1$. Each member of the sequence is rational or irrational. First, we consider the case that there are only finitely many rational numbers. Then $\lim_{n \to \infty}(f(x_n))=f(1)$ is clear. Analogously for the case that there only finitely many irrational numbers.

Hence, we can consider the two subsequence $(x_{1_i})$ and $(x_{2_r})$ which consist only irrational and rational numbers, respectively.

Now, let $\varepsilon>0$ be arbitrary. As we know $f(x_{1_i})\to f(1)$ and $f(x_{2_r})\to f(1)$ there exist $i_0,r_0 \in \mathbb{N}$ such that $$ |f(x_{1_i})-f(1)|<\varepsilon $$ and $$ |f(x_{2_r})-f(1)|<\varepsilon $$ holds for all $i>i_0$ or all $r>r_0$ respectively.

Thus, we can conclude that there exists $n_0 \in \mathbb{N}$ such that $$ |f(x_n)-f(1)|<\varepsilon $$ holds for all $n>n_0$.


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