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Here is Prob. 15, Sec. 5.1, in the book Introduction To Real Analysis by Robert G. Bartle and Donald R. Sherbert, 4th edition:

Let $f \colon (0, 1) \to \mathbb{R}$ be bounded but such that $\lim_{x \to 0} f$ does not exist. Show that there are two sequences $\left(x_n\right)$ and $\left(y_n\right)$ in $(0, 1)$ such that $\lim \left( x_n \right) = 0 = \lim \left( y_n \right)$, but such that $\lim \left( f \left(x_n\right) \right)$ and $\lim \left( f \left(y_n\right) \right)$ exist but are not equal.

My Attempt:

For each $n \in \mathbb{N}$, let us put $$ I_n \colon= \left( 0, \frac{1}{n } \right) = \left\{ \ x \in \mathbb{R} \ \colon \ 0 < x < \frac{1}{n} \ \right\}, \tag{0} $$ and hence let us take $$ \alpha_n \colon= \inf f \left( I_n \right) \qquad \mbox{ and } \qquad \beta_n \colon= \sup f \left( I_n \right). \tag{1} $$ Then as $I_{n+1} \subset I_n$, so we must have $$ \alpha_{n} \leq \alpha_{n+1} \leq \beta_{n+1} \leq \beta_n. \tag{2} $$ And, for any natural numbers $m$ and $n$ such that $m < n$, we see from (2) that $$ \alpha_m \leq \alpha_{m+1} \leq \cdots \leq \alpha_n \leq \beta_n, $$ and also $$ \alpha_n \leq \beta_n \leq \beta_{n-1} \leq \cdots \leq \beta_m. $$ Hence we can conclude without any loss of generality that, for any natural numbers $m$ and $n$, we have $$ \alpha_m \leq \beta_n. \tag{3} $$

Moreover, as our function $f$ is bounded on the open interval $(0, 1)$, so we must also have $$ -\infty < \inf f \big( (0, 1) \big) \leq \sup f \big( (0, 1) \big) < +\infty, \tag{4} $$ and in fact $$ \inf f \big( (0, 1) \big) \leq \alpha_n \leq \beta_n \leq \sup f \big( (0, 1) \big) \tag{5} $$ for all $n \in \mathbb{N}$.

Thus $\left( \alpha_n \right)$ is a monotonically increasing sequence of real numbers which is also bounded from above in $\mathbb{R}$, and $\left( \beta_n \right)$ is a monotonically decreasing sequence of real numbers which is also bounded from below in $\mathbb{R}$. Therefore both of these sequences are convergent in $\mathbb{R}$. Let us put $$ \alpha \colon= \lim_{n \to \infty } \alpha_n \qquad \mbox{ and } \qquad \beta \colon= \lim_{n \to \infty } \beta_n. \tag{6} $$ Then $$ \alpha \leq \beta, \tag{7} $$ by virtue of (3) above.

For each $n \in \mathbb{N}$, as $$ \alpha_n + \frac{1}{n} > \alpha_n \qquad \mbox{ and } \qquad \beta_n - \frac{1}{n} < \beta_n, $$ so (by the definition of the supremum and the infimum of a non-empty bounded subset of $\mathbb{R}$) there exist real numbers $x_n$ and $y_n$ in $I_n$ such that $$ \alpha_n \leq f \left( x_n \right) < \alpha_n + \frac{1}{n} \qquad \mbox{ and } \qquad \beta_n - \frac{1}{n} < f \left( y_n \right) \leq \beta_n. \tag{8} $$ [Refer to (0) and (1) above.]

Thus we have sequences $\left( x_n \right)$ and $\left( y_n \right)$ in the open interval $(0, 1)$ such that $$ 0 < x_n < \frac{1}{n} \qquad \mbox{ and } \qquad 0 < y_n < \frac{1}{n} $$ for all $n \in \mathbb{N}$. Therefore by the sandwiching theorem we can conclude that both of these sequences converge to $0$. But from (6) and (8) together with the sandwiching theorem we can also conclude that
$$ \lim_{n \to \infty} f \left( x_n \right) = \alpha \qquad \mbox{ and } \qquad \lim_{n \to \infty} f \left( y_n \right) = \beta. $$

Is what I've done so far correct? If so, then how to show that our $\alpha$ and $\beta$ in (6) above are different?

Or, have I made a mistake anywhere in my reasoning?

Or, is there any other (and easier and more direct) way of proving this result?

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  • $\begingroup$ Axiom of Choice has been used here, when you choose the sequences x_n and y_n $\endgroup$ Mar 15, 2023 at 13:35
  • $\begingroup$ Although in a lot of Problems in Bartle and Sherbert, it is, and in fact the author has freely used it In examples and in proofs of theorems without mention. So you needn't worry about it if you're following this book $\endgroup$ Mar 15, 2023 at 13:37

1 Answer 1

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What you did is correct.

In order to show that $\alpha\neq\beta$, suppose otherwise. That is, suppose that $\alpha=\beta$. I will prove that $\lim_{x\to0}f(x)=\alpha(=\beta)$, thereby reaching a contradiction. Take $\varepsilon>0$. Now, take $n\in\mathbb N$ such that $\beta_n-\alpha_n<\varepsilon$; it must exist, since $\lim_{n\to\infty}\beta_n-\alpha_n=\beta-\alpha=0$. But then, by the definition of $\alpha_n$ and $\beta_n$,$$x\in\left(0,\frac1n\right)\implies\alpha_n\leqslant f(x)\leqslant\beta_n\implies\bigl|f(x)-\alpha\bigr|<\varepsilon,$$since both $f(x)$ and $\alpha$ belong to $(\alpha_n,\beta_n)$ and $\beta_n-\alpha_n<\varepsilon$.

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