Homeomorphism

**namelessguy** · March 2nd 2008, 09:56 AM

Can someone help me to show that Q is not homeomorphic to N.

We know that there exists a bijection between Q and N. How can I show that f:Q onto N is not continuous ?

I also have trouble with showing any isometry is continuous and is a homeomorphism.

Part 1)
We know that f: M onto N is an isometry if d_N (fp, fq) = d_M (p,q) for each p,q in M. So, f preserves distance. We now want to show that f is continuous. Assume that f is not continuous, then there exists x in M such that f is not continuous at x. This implies that there exists epsilon greater than 0 st for each delta greater than 0 and for each y in M, we have d_M (x,y) less than delta implies d_N (f(x), f(y)) greater than or equal to 0.
I am not able to see a contradiction here. Anyone can help pls?

Part 2)
We want to show that f is a bijection, but this is given in the definition of isometry. We need to show that f is continuous, this will be given if I can prove part 1). The last proof is to prove that inverse of f is continuous. Anyone can help?

**JaneBennet** · March 2nd 2008, 10:26 AM

There is no need to use contradiction for Part (1). It’s totally straightforward.

Let $x\in M$ . You want to show that f is continuous at x. Given $\epsilon>0$ , simply set $\delta=\epsilon$ and you have that $\forall\,y\in M$ , $\mathrm{d}_M(x,y)<\delta\ \Rightarrow\ \mathrm{d}_N(\mathrm{f}(x),\mathrm{f}(y))=\mathrm{ d}_M(x,y)<\delta=\epsilon$ .

Part (2):
f is a function from M onto N; in other words, it is surjective.

Let $x_1,x_2\in M$ . Then

$\mathrm{f}(x_1)=\mathrm{f}(x_2)\ \Rightarrow\ \mathrm{d}_N(\mathrm{f}(x_1),\mathrm{f}(x_2))=0$
$\color{white}.\hspace{22mm}.$ $\Rightarrow\ \mathrm{d}_M(x_1,x_2)=0$
$\color{white}.\hspace{22mm}.$ $\Rightarrow\ x_1=x_2$

So f is injective as well. Hence it is a homeomorphism.

**namelessguy** · March 2nd 2008, 12:52 PM

Originally Posted by JaneBennet

There is no need to use contradiction for Part (1). It’s totally straightforward.

Let $x\in M$ . You want to show that f is continuous at x. Given $\epsilon>0$ , simply set $\delta=\epsilon$ and you have that $\forall\,y\in M$ , $\mathrm{d}_M(x,y)<\delta\ \Rightarrow\ \mathrm{d}_N(\mathrm{f}(x),\mathrm{f}(y))=\mathrm{ d}_M(x,y)<\delta=\epsilon$ .

Part (2):
f is a function from M onto N; in other words, it is surjective.

Let $x_1,x_2\in M$ . Then

$\mathrm{f}(x_1)=\mathrm{f}(x_2)\ \Rightarrow\ \mathrm{d}_N(\mathrm{f}(x_1),\mathrm{f}(x_2))=0$
$\color{white}.\hspace{22mm}.$ $\Rightarrow\ \mathrm{d}_M(x_1,x_2)=0$
$\color{white}.\hspace{22mm}.$ $\Rightarrow\ x_1=x_2$

So f is injective as well. Hence it is a homeomorphism.

Thanks a lot for your help, but I don't know why in part 2) showing f is bijective, then it is homemorphism. Since from what I learned, to show f is homeomorphism, I need to check 3 conditions: f is a bijection, f is continuous, and inverse of f is continuous.
However, I now see how to approach part 2) . Since f is a bijection, the inverse of f is also a bijection, and inverse of f also preserves distance. Hence, inverse of f is isometry. By applying part 1), inverse of f is continuous

**JaneBennet** · March 2nd 2008, 05:38 PM

You’re right, one has to check that $\mathrm{f}^{-1}$ is continuous as well.

Since f is a bijection, we can write $N=\{\mathrm{f}(x):x\in M\}$ .

Let $\mathrm{f}(x)\in N$ . You want to show that $\mathrm{f}^{-1}$ is continuous at $\mathrm{f}(x)$ . Given $\epsilon>0$ , simply set $\delta=\epsilon$ and you have that $\forall\,\mathrm{f}(y)\in N$ , $\mathrm{d}_N(\mathrm{f}(x),\mathrm{f}(y))<\delta\ \Rightarrow\ \mathrm{d}_M(\mathrm{f}^{-1}(\mathrm{f}(x)),\mathrm{f}^{-1}(\mathrm{f}(y)))=\mathrm{d}_M(x,y)=\mathrm{d}_N( \mathrm{f}(x),\mathrm{f}(y))<\delta=\epsilon$ .

**namelessguy** · March 2nd 2008, 06:16 PM

Do you have any suggestion for me on how to show Q is not homeomorphic to N. Where Q is the set of rationals, and N is the set of naturals.

**JaneBennet** · March 2nd 2008, 06:33 PM

Show that there are no isometries between $\mathbb{Q}$ and $\mathbb{N}$ . Make use of the fact that distances in $\mathbb{Q}$ can be arbitrarily small whereas distances in $\mathbb{N}$ can’t be arbitrarily small. (I mean of course distances between distinct elements.)

**namelessguy** · March 2nd 2008, 08:37 PM

Originally Posted by JaneBennet

Show that there are no isometries between $\mathbb{Q}$ and $\mathbb{N}$ . Make use of the fact that distances in $\mathbb{Q}$ can be arbitrarily small whereas distances in $\mathbb{N}$ can’t be arbitrarily small. (I mean of course distances between distinct elements.)

I thought that isometry is a stronger property than homeomorphism. So, if f is an isometry, then f is a homeomorphism. However, if f is not an isometry, then it doesn't guarantee that f is not a homeomorphism. I think this way because I take the contrapositive statement: if f is not homeomorphism, then f is not isometry.

**JaneBennet** · March 3rd 2008, 01:57 AM

The correct contrapositive statement is: If f is not an isometry, then f is not a homeomorphism.

In any case, f can be an isometry without being a homeomorphism. For example, $\mathrm{f}:\mathbb{N}\to\mathbb{Q},\ \mathrm{f}(n) = n+\mbox{$\frac{1}{2}$}$ . Hence the statement “if f is not a homeomorphism, then f is not an isometry” is false.

Try this: Let $\mathrm{f}:M\to N$ be an isometry, $y\in M$ be a fixed point and $(x_n)_{n\,=\,1}^\infty$ be a sequence in M such that $x_n\ne y$ for any $n\in\mathbb{N}$ . Then if $\mathrm{d}_M(x_n,y)\to0$ as $n\to\infty$ , $\mathrm{d}_N(\mathrm{f}(x_n),\mathrm{f}(y))\to0$ as $n\to\infty$ .

Now suppose $\mathrm{f}:\mathbb{Q}\to\mathbb{N}$ is an isometry. Let $y=0$ and $x_n=\frac{1}{n}$ . We have shown above that any isometry is injective; hence $x_n\ne y\ \Rightarrow\ \mathrm{f}(x_n)\ne\mathrm{f}(y)$ for any n. But $\lim_{n\,\to\,\infty}{\mathrm{d}_\mathbb{Q}(x_n,y) } = 0$ , whereas $\mathrm{d}_\mathbb{N}(\mathrm{f}(x_n),\mathrm{f}(y ))$ can never tend to zero if $\mathrm{f}(x_n)\ne\mathrm{f}(y)$ since the set $\{\mathrm{d}_\mathbb{N}(m,n):m,n\in\mathbb{N},m\ne n\}$ has a lower bound of 1. This is a contradiction – hence f cannot be an isometry.

Thread: Homeomorphism

LinkBack

Thread Tools

Display

Homeomorphism

Similar Math Help Forum Discussions

Homeomorphism

Homeomorphism

homeomorphism between R^(n^2) and M_n(R)

homeomorphism

Homeomorphism

Search Tags