Proving a function is continuous but unbounded

continuityconvergence-divergencereal-analysis

Problem:

Hi, I'm working on the following math problem:

Suppose a set $S$ contains a sequence that converges to a point
$x_{0}$ that is not in $S$. Show that the function $f : S
\rightarrow \mathbb{R}$ defined by $f(x) = 1/|x – x_{0}|$ for all $x$
in $S$ is continuous but unbounded.

Here is the definition of continuity that I am referring to. Note, however, that I am not using the typical $\epsilon-\delta$ definition of continuity:

Definition: A function $f : D \rightarrow \mathbb{R}$ is said to be continuous at the point $x_{0}$ in $D$ provided that whenever
$\{x_{n}\}$ is a sequence in $D$ that converges to $x_{0}$, the image
sequence $\{f(x_{n})\}$ converges to $f(x_{0})$.

My attempt:

Let $\{x_{n}\}$ be a sequence in $S$ that converges to the point $x_{0}$, which is not in $S$. By the definition of convergence provided, this means $\forall \epsilon > 0, \exists N$ s.t.

$$|x_{n} – x_{0}| < \epsilon$$

$\forall n \geq N$. To prove continuity at $x_{0}$, I must show that there is some index $N'$ so that $\forall \epsilon > 0$,

$$|f(x_{n}) – f(x_{0}) | < \epsilon$$

$\forall \epsilon > 0$. By our definition of $f$, I can write

$$|f(x_{n}) – f(x_{0}) | = |1/(x – x_{0}) – 1/(x_{0} – x_{0})|,$$

but the right-hand term is not defined, so I'm not sure if my approach is correct.

Some additional info:

In case it helps, this problem is part of a two-part problem. The first part of this problem asked me to show that if $S$ contains an unbounded sequence, then show that the function $f : S \rightarrow \mathbb{R}$ defined by $f(x) = x$ for all $x$ in $S$ is continuous but unbounded. So, I can cite this fact if I want. I'm guessing that I'll have to since it's part of the same problem, but I'm not sure where that'll come into play.

Thank you to anyone who tries to help.

Best Answer

Your approach is not correct because $x_0\notin D$. So, it makes no sense to ask whethr $f$ is continuous at $x_0$. The function $f$ is continuous because, for instance:

  • the function $x$ is continuous;
  • so, $x-x_0$ is continuous;
  • since the absolute value function is continuous, $\lvert x-x_0\rvert$ is continuous;
  • therefore, $\dfrac1{\lvert x-x_0\rvert}$ is continuous.

And $f$ is unbounded because, if $(x_n)_{n\in\mathbb N}$ is such that $\lim_{n\to\infty}x_n=x_0$, then $\lim_{n\to\infty}f(x_n)=+\infty$.

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