Compactness under different metric? 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?
 A: 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.
A: 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.
A: 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 $y$.
