Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
What is/are your question(s) then? –  Ittay Weiss Feb 25 '13 at 19:47
    
I need help for i), ii) and iii) –  aa_x Feb 25 '13 at 19:51
    
do you know the definitions of the concepts involved? –  Ittay Weiss Feb 25 '13 at 19:55
    
i) the definition I remember is d(x,y)=> x=y d(x,y)=d(y,x)and the triangular inequality. I don't know which other definitions which are usefull to this exercise –  aa_x Feb 25 '13 at 20:12
1  
i) Can you show $\rho(x,x)=0$? Can you show that $\rho(x,y)>0$ if $x\ne y$? Can you show $\rho(x,y)=\rho(y,x)$? And the triangle inequality? Use that $d$ is a metric and basic properties of the absolute value. - By the way, why are we allowed to divide by $d(x,X\setminus U)$ in the first place? –  Hagen von Eitzen Feb 25 '13 at 20:36

1 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: $$\left|\frac{1}{d(x,X\setminus U)}-\frac{1}{d(z,X\setminus U)}\right|=\left|\frac{1}{d(x,X\setminus U)}-\frac{1}{d(y,X\setminus U)}+\frac{1}{d(y,X\setminus U)}-\frac{1}{d(z,X\setminus U)}\right|$$


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.


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 $d$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.