Topology - axioms of metric space - convergency - Cauchy

Question

Let $(X,d)$ be a complete metric space and $U \subseteq X$, $U \neq X$, its open subset. Define a function $\rho\colon U \times U \rightarrow [0, \infty)$ as: $$\rho(x,y):=d(x,y)+\left|\frac{1}{d(x,X\setminus U)} - \frac{1}{d(y,X\setminus U)}\right|,$$ where $d(x,X\setminus U)$ is the usual distance between point $x$ and subset $X\setminus U$:

i) Show that function $\rho$ satisfies the axioms of a metric.

ii) Let $(x_n)$ be a sequence in $U$ and $w \in U$. Show that the sequence $(x_n)$ converges to $w$ in metric $d$ if and only if it converges to $w$ in metric $\rho$. Thus, the two metrics $d$ and $\rho$ give rise to the same topology on $U$.

iii) Let $(x_n)$ be a sequence in $U$, which is Cauchy with respect to metric $\rho$. Show that $(x_n)$ is also Cauchy with respect to metric $d$, and thus it converges to some point $y \in X$. Show that $y \in U$, since otherwise the sequence $(x_n)$ would be unbounded with respect to metric $\rho$. Conclude that $(U,\rho)$ is a complete metric space.

I need help for i), ii) and iii). I don't know how to solve them.

Answer 1

For (i), you need to show the following:

(a) $\rho(x,x)=0$ for any $x\in U$ (plug it in, and use metric properties of $d$), and $\rho(x,y)>0$ for $x,y\in U$ with $x\neq y$ (plug it in, use metric properties of $d$, and use the fact that the absolute value expression is at least $0$).

(b) $\rho(x,y)=\rho(y,x)$ for $x,y\in U$ (use metric properties of $d$ and absolute value properties).

(c) $\rho(x,z)\leq\rho(x,y)+\rho(y,z)$ for $x,y,z\in U$. For that, use metric properties of $d$, absolute value properties, and the following observation: $$\left|\frac{1}{d(x,X\setminus U)}-\frac{1}{d(z,X\setminus U)}\right|=\left|\frac{1}{d(x,X\setminus U)}-\frac{1}{d(y,X\setminus U)}+\frac{1}{d(y,X\setminus U)}-\frac{1}{d(z,X\setminus U)}\right|$$

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For (ii), notice first of all that $\rho(x,y)\geq d(x,y)$ for all $x,y\in U$. Hence, if $\rho(x_n,w)\to 0$, then $d(x_n,w)\to 0$. To show that the other implication holds, it suffices (why?) to show that if $d(x_n,w)\to 0$, then $$\left|\frac{1}{d(x_n,X\smallsetminus U)}-\frac{1}{d(w,X\smallsetminus U)}\right|\to 0.$$ You should really use $\epsilon$-$n$ definitions of convergence, here, but that's the general idea.

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For (iii), recall (again) that $\rho(x,y)\geq d(x,y)$ for all $x,y\in U$, and use that to show that a Cauchy sequence (with respect to the metric $\rho$) of points in $U$ will be Cauchy with respect to $d$. Remember that a metric space is complete if and only if Cauchy sequences of points in the space converge in the space. It remains to show (as noted) that if a sequence of points in $U$ converges to a point in $X\setminus U$, then the sequence is unbounded with respect to the metric $\rho$, and so can't be Cauchy with respect to $\rho$ (since Cauchy sequences are bounded), even though it is Cauchy with respect to $d$.