# how to find a metric to make a space complete (help)

Hi everyone I'm struggle with the following.

Define a complete metric on $\mathbb{R}\setminus \{0,1\}$ with usual relative topology. I'd like to follow the big hint of Daniel Fischer but I have troubles because this is new:

Take a continuous function $f\colon \mathbb{R}\setminus \{0,1\}\to\mathbb{R}$, and consider $\Gamma(f) = \left\{ (t,f(t)) : t\in \mathbb{R}\setminus\{0,1\}\right\}$. The graph $\Gamma(f)$ inherits a metric from $\mathbb{R}^2$, and transporting that back, you get a metric on $\mathbb{R}\setminus\{0,1\}$ inducing the standard topology. Now you need to choose $f$ so that $\Gamma(f)$ is a closed subset of $\mathbb{R}^2$.

Suppose I have that continuous map, then $g:\mathbb{R}\setminus \{0,1\}\to \Gamma(f)$ defined by $t\to (t,f(t))$ is continuous, the $\text{pr}_1$ is its inverse which is continuous. But from here I have several questions:

1) How to define the metric on $\mathbb{R}\setminus \{0,1\}$, I don't know my guess is something like $d(x,y)=d_E(g(x),g(y))$ where $d_E$ is the euclidean metric inherited to $\Gamma(f)$

2) How this shows that $\mathbb{R}\setminus \{0,1\}$ is complete I know that $\Gamma(f)$ is because I'm assuming that is closed and how I know that it has the same topology as $\Gamma(f)$ because in this form the function which I defined as the metric is not the correct one, I think.

3) Some reference to study this particular kind of argument because are totally new for me and very interesting (I suppose any book in general topology but I prefer to know something more specific).

EDIT

Thanks in advance and I'm apologize if my questions are very stupid but this way of thinking is completely new for me. My classes of topology was one introductory courses in Real Analysis.

4) From here how do we know that $d$ as above using for example $x\to 1/x(x-1)$ give us the same topology as $d\restriction \mathbb{R}\setminus \{0,1\}\times \mathbb{R}\setminus \{0,1\}$

• The problem, of course, with the standard metric on $\mathbb{R}-\{0,1\}$ re: completeness is that you can have Cauchy sequences approaching $0$ or $1$. Your proposed metric $d_E(g(x),g(y))$ is good, but $f$ needs to be such that $d_E$ avoids these problematic Cauchy sequences...say, a function with vertical asymptotes at $0$ and $1$. – Nick D. Sep 12 '14 at 0:36
• @NickD. This is to avoid that the cauchy sequences which approaches to $1$ or $0$ can be Cauchy sequences in our new metric? this is the general idea right? – Jose Antonio Sep 12 '14 at 0:41
• Right, say if you choose $f(x)=1/(x^2-x)$, what is $d(.001,.0001)$ in the metric you're defining on $\mathbb{R}-\{0,1\}$? This should give the intuition that sequences "approaching" $0$ are not Cauchy. – Nick D. Sep 12 '14 at 0:50
• @NickD. my first guess was $\frac{2t-1}{t(t-1)}$ by simplicity $1/x+1/(x-1)$ but yours is nicer. So only I need to check that the set $\Gamma(f)$ is closed in $R^2$ and with this only check that the completness follows from the completness inherited to the set $\Gamma(f)$ (if is closed) after this I have to check that generates the same topology as the relative topology with the usual metric, right? in this case what could be a good approach try to show that the metric has to be equivalent or something like that ? Sorry soryy is the first time that I see this kind of problems – Jose Antonio Sep 12 '14 at 0:53
• @NickD. On more thing, to show that generates the same topology as the relative. We can take any open in $\mathbb{R}\setminus\{0,1\}$ say $A$. Thus $B=A\times \mathbb{R}$ has to be open in $\mathbb{R}$ and for instances $B\cap \Gamma$ is also open. And from here to show that $A$ is open in the new metric is suffice to show that the new metric $A$ correspond to $B\cap \Gamma$, I'm not sure – Jose Antonio Sep 12 '14 at 13:11

(1) That is the right guess, now you just have to choose the right $g$. You want some $g$ that makes $0$ and $1$ infinitely far away, so that you can't have a cauchy sequence approaching $0$ or $1$. This should give you a hint on how the graph looks.
(2) It shows that $\mathbb{R} \backslash \{0,1\}$ is complete because it is isometric to $\Gamma(f)$ with euclidean metric, and $\Gamma(f)$ is complete because closed subspaces of complete spaces are complete.
• Thinking quikly, I don't maybe $f$ may be something like $1/x+1/(x-1)$ everything close to zero and to one goes in opposite direction so the cauchy sequences in R wiht the usual topology doesnt have chance to be Cauchy if the approaches 1 or 0 – Jose Antonio Sep 12 '14 at 0:46
• only I need to show that $\Gamma(f)$ is closed in $R^2$ – Jose Antonio Sep 12 '14 at 0:47
• tentatively I can use $t\mapsto \frac{2t-1}{t(t-1)}$ and I have to show that $\Gamma(f)= \{(x,f(t))\}$ is closed – Jose Antonio Sep 12 '14 at 0:51
• @I have a problem with one point how to show that induce the same topology in $\mathbb{R}\\{0,1\}$ with the usual. – Jose Antonio Sep 12 '14 at 3:28