Compactness under different metric?

Question

Consider the metric $\rho(x,y)=\frac{|x-y|}{1+|x-y|}$ on $\mathbb{R}$. Is $(\mathbb{R},\rho)$ compact?

In order to show that is not, I wanted to find a sequence such that any subsequence is non-convergent and I considered $x_n=n$. Does this work or I missed something?

Answer 1

An alternative approach to your problem:

First, show that the function $f:(\Bbb R,\rho) \to (\Bbb R, |\cdot|)$ taking $x \mapsto x$ is continuous. Now, the continuous image of a compact set is compact.

So, suppose that $\Bbb R$ were compact under $\rho$. Then since $f$ is continuous, $f(\Bbb R) = \Bbb R$ would be compact under $|\cdot|$. Since this is not true, we have derived a contradiction.

Answer 2

Indeed, the sequence defined by $x_n = n$ works. To see this, let $y_n = x_{\phi(n)}$ be a subsequence. Let $y\in\mathbb R$. We have $$\begin{align} \lim_{n\rightarrow\infty} \frac{|y_n-y|}{1+|y_n-y|} &= \lim_{n\rightarrow\infty} \frac{|\phi(n)-y|}{1+|\phi(n)-y|}\\ &= \lim_{n\rightarrow\infty} \frac{|\phi(n)-y|}{1+|\phi(n)-y|}\times\frac{1+\phi(n)}{1+\phi(n)}\times\frac{\phi(n)}{\phi(n)}\\ &= \lim_{n\rightarrow\infty} \frac{|\phi(n)-y|}{\phi(n)}\times\frac{1+\phi(n)}{1+|\phi(n)-y|}\times\frac{\phi(n)}{1+\phi(n)} \\ &= \lim_{n\rightarrow +\infty} \frac{\phi(n)}{1+\phi(n)}=1 \end{align}$$ so that $(y_n)$ doesn't converge to $y$.

Answer 3

Every metric space $(M,d)$ determines the associated topological space $(M,\mathcal{O})$ whose open sets $U\in\mathcal{O}$ are the unions of open balls of the metric space. Compactness of a metric space is a topological property, meaning that the metric space is compact iff the associated topological space is compact. Two homeomorphic topological spaces have the same topological properties; in particular, they are either both compact or neither is compact.

In your case let $d(x,y)=|x-y|$ be the usual metric on $\mathbb{R}$. Then the identity function $\mathbb{R}\to\mathbb{R} : x\mapsto x$ is a uniformly continuous mapping from the metric space $(\mathbb{R},d)$ to the metric space $(\mathbb{R},\rho)$, while its inverse, which is also the identity function, is a uniformly continuous mapping from the metric space $(\mathbb{R},\rho)$ to the metric space $(\mathbb{R},d)$. It follows that the topological spaces associated with the two metrics on $\mathbb{R}$ are homeomorphic, and since this particular homeomorphism is the identity on $\mathbb{R}$, the two topological spaces are in fact identical. In short, the metric $\rho$ determines the same topology on $\mathbb{R}$ as the metric $d$, which is the topology of the real line. Now the answer to your question is obvious, is it not?

By the way, instead of the metric $\varrho$ you can choose the metric $\alpha(x,y)=\min(|x-y|,a)$, where $a$ is a (fixed) strictly positive real number, say $a=10^{-1000}$. As above, the identity mapping is uniformly continuous from the metric space $(\mathbb{R},d)$ to the metric space $(\mathbb{R},\alpha)$, and the same is true in the opposite direction, so the topology determined by the metric $\alpha$ is the usual topology of the real line.