$(0,1)$ is compact w.r.t a metric which induces the usual topology on $\Bbb R$.

Which of the following is true?

1. $(0,1)$ is complete w.r.t a metric which induces the usual topology on $\Bbb R$.

2. $(0,1)$ is compact w.r.t a metric which induces the usual topology on $\Bbb R$.

1. True. Since $(0,1)$ and $\Bbb R$ have the same cardinality hence there exists a bijective function $f:(0,1)\to \Bbb R$. Let's define a metric $d_1$ on $(0,1)$ by $d_1(x,y)=\left|f(x)-f(y)\right|$ which defines a isometry.

Now since $\Bbb R$ is complete so will be $(0,1)$.

1. False since in order to be compact $(0,1)$ must be closed in the usual topology which will never be.

• Hmmm. I can see your reasoning for #1, but then again there are Cauchy sequences in $(0, 1)$ that don't converge - e.g. $a_n = 1-1/n$. I don't know if I'm correct in my thinking, but if so, it's interesting that completeness isn't preserved under homeomorphism. You are correct about #2. In addition to appealing to Heine-Borel, one can also note that the open cover $(1/n, 1 \! - \! 1/n)$ has no finite subcover. – Kaj Hansen Dec 10 '16 at 5:44
• I would assume that in these questions, the metric is supposed to be a metric on all of $\mathbb{R}$, not just a metric on $(0,1)$ (otherwise, what does "which induces the usual topology on $\mathbb{R}$" mean?). – Eric Wofsey Dec 10 '16 at 5:46