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I have tried unsuccessfully to follow a hint in Gamelin & Greene's Topology, 2nd Edition, Dover, ex.9 in Section 6, that is written to demonstrate the existence of a continuous, real valued, continuous but unbounded function on a metric space that is not complete and therefore not compact. The author of the hint uses the real metric to create a function from it using a Cauchy sequence that does not converge. I don't want to detail the hint, because there are frequent typos in the book, most regrettably, and I don't know if I reading it as the author intended. The author does seem to show that the inverse of the function is unbounded and continuous using the Cauchy sequence.

Does anyone know such a construction, using the inverse of d($x,x_n$) ?

goedelite
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1 Answers1

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Consider a metric space $M$ that is not complete. Let $\{x_n\}_{n \in \mathbb{N}}$ be a Cauchy sequence in $M$ with no limit point.

Let $\hat{M}$ be the Cauchy completion of $M$, with the inclusion map $z \mapsto \hat{z} : M \to \hat{M}$. Let $w$ be the limit point $w = \lim\limits_{n \to \infty} \hat{x_n}$.

Define a function $f : M \to \mathbb{R}$ by $f(x) = \frac{1}{d(\hat{x}, w)}$. This is a continuous function since $x \mapsto \hat{x}$ is continuous, $y \mapsto d(y, w)$ is continuous, and $z \mapsto \frac{1}{z}$ is continuous.

However, $f$ is not bounded because $\lim\limits_{n \to \infty} d(\hat{x_n}, w) = d(\lim\limits_{n \to \infty} \hat{x_n}, w) = d(w, w) = 0$, and therefore $\lim\limits_{n \to \infty} \frac{1}{d(\hat{x_n}, w)} = \infty$.

If we dig into the precise definition of the completion of a metric space, we see that $f(x) = \frac{1}{\lim\limits_{n \to \infty} d(x, x_n)}$. Proving that $f$ is a continuous function from this definition directly is a bit more challenging, but it can be done.

Edit because of comments: the limit $\lim\limits_{n \to \infty} d(x, x_n)$ exists for all $x \in M$. Note that by the triangle inequality, we have $|d(x, x_n) - d(x, x_m)| \leq d(x_n, x_m)$. Consider an arbitrary $\epsilon > 0$. Take an $N$ such that for all $m, n \geq N$, $d(x_n, x_m) < \epsilon$. Then for all $m, n \geq N$, $|d(x, x_n) - d(x, x_m)| < \epsilon$. Thus, the sequence $d(x, x_n)$ is a Cauchy sequence in $\mathbb{R}$, hence it has a limit. The limit cannot be zero, or $x$ would be the limit of $x_n$.

It can be shown that $\lim\limits_{n \to \infty} d(x, x_n)$ is continuous with respect to $x$. In fact, each function $x \mapsto d(x, x_n)$ is continuous with respect to $x$, and the sequence uniformly converges in $x$ to $\lim\limits_{n \to \infty} d(x, x_n)$.

This demonstrates the alternate definition of $f$ is a continuous function. One can again show that $\lim\limits_{m \to \infty} \frac{1}{\lim\limits_{n \to \infty} d(x_m, x_n)} = \infty$.

Mark Saving
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  • Mark, your last sentence contains "$\lim d(x,x_n)$". But that quantity has no limit because ${x_n}$ is a non-convergent Cauchy sequence. Is the little roof symbol missing that is present in the preceding sentence? Otherwise, I follow you. – goedelite May 02 '21 at 23:17
  • @goedelite It turns out that $\lim d(x, x_n)$ actually does exist as long as $x$ is a Cauchy sequence, but showing it requires a bit of work. Basically, you'll want to show that the sequence $a_n = d(x, x_n)$ is itself a Cauchy sequence of real numbers. Then $a_n$ will have a limit. You can use $|a_n - a_m| = |d(x, x_n) - d(x, x_m)| \leq d(x_n, x_m)$ to prove that $a$ is a Cauchy sequence. – Mark Saving May 02 '21 at 23:43
  • I have no doubt that the limit exists if $x_n$ is a C.S. I commented because it is not a C.S. in the argument. Thus the limit with an x with a chapeau exists in the completion, as you presented, but not otherwise, right? – goedelite May 03 '21 at 00:16
  • @goedelite I've filled in the argument to fully explain why the second definition of $f$ is a valid definition. – Mark Saving May 03 '21 at 23:51