# More Metric Spaces

• Nov 19th 2007, 07:53 AM
Jhevon
More Metric Spaces
Here are two more questions that have been bothering me. for the first, i don't think my solution is good enough, there's somethign too "obvious" about it. for the second, i haven't a clue

1) Let $X$ and $Y$ be metric spaces. Assume that $X$ is complete and that there exists $f: X \to Y$ which is bijective and continuous. Suppose that $f^{-1}$ is uniformly continuous. Prove that $Y$ is complete

2) Prove that any open subset of a complete metric space is homeomorphic to a complete metric space. (Hint: If $A$ is the complement of the open set, use the metric
$\left. D_1(x,y) = D(x,y) + \left| \frac 1{D(x,A)} - \frac 1{D(y,A)} \right| \right)$

Here are definitions/theorems i figured would come in handy:

Defintion: A metric space $M$ is complete if every Cauchy sequence in $M$ is convergent to a point of $M$.

Theorem: Any convergent sequence in a metric space is a Cauchy sequence.

Definition: A function $f$ from a metric space $X$ to a metric space $Y$ is uniformly continuous if for every $\epsilon > 0$ there exists a $\delta > 0$ such that, for all $x,x' \in X$, $D(x,x') < \delta$ implies $D[f(x),f'(x)] < \epsilon$.

Theorem: A uniformly continuous function carries Cauchy sequences into Cauchy sequences.

Definition: Let $f_i,f$ be functions from a metric space $X$ to a metric space $Y$. We say that the sequence $f_i$ converges uniformly to $f$ if the following is true: For any $\epsilon > 0$ there exists $N$ (depending on $\epsilon$ but independent of $x$) such that $D[f_i(x),f(x)]< \epsilon$ for all $i \ge N$ and all $x \in X$.

Definition: Two metric spaces are said to be homeomorphic if there exists between them a bijection which is continuous in both directions.

Whew! ok, here's what i did.

...oh, wait, i have to go to class. I'll post my "proof" to question 1) later, there's not enough time.

for question 2)

Proof:

I have no clue, but I have a gut feeling it is true :D (they wouldn't ask me to prove something that was false, after all)

QED

Ok, so of course that proof was a joke, and a bad one, i know. but i'm not sure what to do
• Nov 20th 2007, 04:57 AM
Opalg
Quote:

Originally Posted by Jhevon
1) Let $X$ and $Y$ be metric spaces. Assume that $X$ is complete and that there exists $f: X \to Y$ which is bijective and continuous. Suppose that $f^{-1}$ is uniformly continuous. Prove that $Y$ is complete

If $(y_n)$ is a Cauchy sequence in Y, then the uniform continuity of $f^{-1}$ implies that $(f^{-1}(y_n))$ is Cauchy in X, and therefore converges to some point $x_0\in X$. The continuity of f then tells you that $y_n\to f(x_0)$. So Y is complete.
Quote:

Originally Posted by Jhevon
2) Prove that any open subset of a complete metric space is homeomorphic to a complete metric space. (Hint: If $A$ is the complement of the open set, use the metric $D_1(x,y) = D(x,y) + \left| \frac 1{D(x,A)} - \frac 1{D(y,A)} \right|$ .)

Call the space X, and the open subset Y. There are two things to prove here. First, that the identity map from Y with the metric D to Y with the metric D_1 is a homeomorphism. Second, that D_1 is a complete metric on Y.

Since $D(x,y)\leqslant D_1(x,y)$, it is clear that the identity map is continuous from (Y,D_1) to (Y,D). For the other direction, suppose that $x_n\to y$ in (Y,D). Then $D(x_n,A)\to D(y,A)$, and it follows that $D_1(x_n,y)\to0$. That is (an outline of) the proof that the identity is a homeomorphism.

To show that (Y,D_1) is complete, let $(x_n)$ be a Cauchy sequence in (Y,D_1). Then $(x_n)$ is Cauchy in (Y,D) and so has a limit $x_0\in X$. You have to show that $x_0\in Y$. For this, notice that $\left| \frac 1{D(x_m,A)} - \frac 1{D(x_n,A)} \right|\leqslant D_1(x_m,x_n)$, and so $\left(1/{D(x_n,A)}\right)$ is a Cauchy sequence of real numbers. So it converges to a limit c. Therefore $D(x_n,A)\to1/c\neq0$. Hence $D(x_0,A) = 1/c>0$, which shows that $x_0\in Y$.
• Nov 20th 2007, 10:00 AM
Jhevon
Quote:

Originally Posted by Opalg
If $(y_n)$ is a Cauchy sequence in Y, then the uniform continuity of $f^{-1}$ implies that $(f^{-1}(y_n))$ is Cauchy in X, and therefore converges to some point $x_0\in X$. The continuity of f then tells you that $y_n\to f(x_0)$. So Y is complete.

my solution was very close to that. somehow i didn't see why uniform convergence was needed, but now i do. thanks

Quote:

Call the space X, and the open subset Y. There are two things to prove here. First, that the identity map from Y with the metric D to Y with the metric D_1 is a homeomorphism. Second, that D_1 is a complete metric on Y.

Since $D(x,y)\leqslant D_1(x,y)$, it is clear that the identity map is continuous from (Y,D_1) to (Y,D). For the other direction, suppose that $x_n\to y$ in (Y,D). Then $D(x_n,A)\to D(y,A)$, and it follows that $D_1(x_n,y)\to0$. That is (an outline of) the proof that the identity is a homeomorphism.

To show that (Y,D_1) is complete, let $(x_n)$ be a Cauchy sequence in (Y,D_1). Then $(x_n)$ is Cauchy in (Y,D) and so has a limit $x_0\in X$. You have to show that $x_0\in Y$. For this, notice that $\left| \frac 1{D(x_m,A)} - \frac 1{D(x_n,A)} \right|\leqslant D_1(x_m,x_n)$, and so $\left(1/{D(x_n,A)}\right)$ is a Cauchy sequence of real numbers. So it converges to a limit c. Therefore $D(x_n,A)\to1/c\neq0$. Hence $D(x_0,A) = 1/c>0$, which shows that $x_0\in Y$.