1. ## Homeomorphism&Embedding

In the first part of the question I've proved that :
The boundary of the square that its vertex are: (0,1) , (1,1), (1,0), (0,0)
is homeomorphic to the unit circle (x^2+y^2=1) as sub-spaces of R^2.

Now I need to prove that the unit circle (as a sub-space of R^2) is not homeomorphic to [0,1] as a sub-space of Rs (Sorgenfrey's line...)

HELP IS NEEDED !

TNX a lot!

2. Originally Posted by WannaBe
In the first part of the question I've proved that :
The boundary of the square that its vertex are: (0,1) , (1,1), (1,0), (0,0)
is homeomorphic to the unit circle (x^2+y^2=1) as sub-spaces of R^2.

Now I need to prove that the unit circle (as a sub-space of R^2) is not homeomorphic to [0,1] as a sub-space of Rs (Sorgenfrey's line...)

HELP IS NEEDED !

TNX a lot!

Hint 1: $\displaystyle S^1\subset \mathbb{R}^2$ is compact, but $\displaystyle [0,1]\subset \mathbb{R}_\mathfrak{S}$ is not

Hint 2: if $\displaystyle f:S^1\rightarrow [0,1]$ is a homeomorphism, then $\displaystyle \left\{f^{-1}\left(\displaystyle{\frac{n}{n+1}}\right)\right\ }_{n=1}^\infty$ is an infinite sequence in $\displaystyle S^1$ and thus it has a convergent subsequence $\displaystyle \Longrightarrow \left\{\displaystyle{\frac{n}{n+1}}\right\}=f\left (f^{-1}\left(\displaystyle{\frac{n}{n+1}}\right)\right) _{n=1}^\infty$ also has a convergent subseq. in [0,1]...but it doesn't, since a sequence converges in the Sorgenfrey Line only if it converges in the euclidean

line FROM ABOVE, meaning: the sequence converges to a finite limit and there's only a finite number of elements of the seq. BELOW that limit...

Tonio

3. Hey tonio....
I'm sry but we didn't learn about compact groups yet...And I didn't quite understand your second hint...I don't think we've learned it... The answer should be pretty much trivial...something about topological qualities...

Hope you'll be able to help me under these restrictions...

4. Originally Posted by WannaBe
Hey tonio....
I'm sry but we didn't learn about compact groups yet...And I didn't quite understand your second hint...I don't think we've learned it... The answer should be pretty much trivial...something about topological qualities...

Hope you'll be able to help me under these restrictions...

Well, it doesn't look very logical that you people deal with homeomorphisms of topological spaces but you haven't yet studied compacticity....there's another hint, but I'm afraid it is more advanced than the previous ones: $\displaystyle S^1\subset\mathbb{R}^2$ is second contable, and thus the image of any continuous open function (like a homeomorphism, say) from it to some top. space is again 2nd. contable...but $\displaystyle \mathbb{R}_\mathfrak{S}$ isn't 2nd contable...and I really can't find anything as trivial as compacticity or limits of sequences. Sorry.

Tonio

5. Hey there tonio...
I know what 2nd countability is... But why you've said S1 is 2nd countable?

TNX

6. Originally Posted by WannaBe
Hey there tonio...
I know what 2nd countability is... But why you've said S1 is 2nd countable?

TNX

Because $\displaystyle \mathbb{R}^2$ is 2nd countable (it has a countable basis for its (euclidean) topology, and thus any subspace of it has the same countable basis for it with the inherited topology)...and again: how come you already know this but not what compacticity is? What book do you guys read in class? Perhaps I am confusing the order of things and compacticity is taught later? Hmm...

Tonio

7. LOL but how can we tell that the subspace of RSis also 2nd countavle? (I mean [0,1] as a subspace)...?

8. Originally Posted by WannaBe
I need to prove that the unit circle (as a sub-space of R^2) is not homeomorphic to [0,1] as a sub-space of Rs (Sorgenfrey's line...)
Have you come across the concept of connectedness? The unit circle is connected. An interval in the S-line is not.

9. Nope...We didn't learn connetedness yet...And 2nd countability is also very vague because we've just said it's a topological quality and that's it....

I'll be delighted to get more help
TNX

10. Originally Posted by WannaBe
LOL but how can we tell that the subspace of RSis also 2nd countavle? (I mean [0,1] as a subspace)...?
Well, it isn't (with the inherited topology of $\displaystyle \mathbb{R}_\mathfrak{S}$) ...for a quick proof:

let $\displaystyle B$ be a basis for the inherited topology of $\displaystyle [0,1]\Longrightarrow \forall r\in [0,1]\,\,\,and\,\,\,\forall\,\,neighborhood\,\,U_r$ of $\displaystyle r\,\,\,\exists\,V_r\in B\,\,\,s.t.\,\,\,r \in V_r\subset U_r$.

Let us then choose $\displaystyle U_r:=[r,1)\subset[0,1]\,,\,\,\forall\,\,r\in [0,1]$ . Suppose now that for $\displaystyle r,s\in [0,1]$ we have $\displaystyle V_r=V_s$ : it is easy to show that $\displaystyle r=s$ , and then we get an injection $\displaystyle V_r \rightarrow r$ from $\displaystyle B$ to $\displaystyle [0,1]$ ...

Tonio