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Let $\mathbb{R}^n$ equipped with the Euclidean topology. Identifying the point $(x_1,\ldots,x_n)$ to $(x_1,\ldots,x_n,0)$ of $R^{n+1}$, we can define the "direct limit" $\mathbb{R}^\infty$ equipped with the direct limit topology i.e. $U \subseteq \mathbb{R}^\infty$ is open iff $U \cap \mathbb{R}^n \subseteq \mathbb{R}^n$ is open.

Denote $(b_n)_{k\in \Bbb{N}}$ to be the canonical basis of $\mathbb{R}^\infty$ so that the canonical basis of $\mathbb{R}^n$ is $(b_1,\ldots,b_n$ which can be written as $\sum_{k=1}^n x_nb_k$.

Now define $S^\infty \subseteq \mathbb{R}^\infty$ to be the subset consisting of tuples such that $\sum_i |x_i|^2 = 1$, equipped with the subspace topology and $H^{\infty}$ be the hyperplane of equation $x_1=0.$

I would like to prove that $S^{\infty}\setminus \{b_1\}$ is homeomorphic to $H^{\infty}$.

I was thinking about $f: (x_2,\ldots,x_n)\mapsto (0,x_2,\ldots,x_n)$. But I think I misunderstood the question.

EDIT: Perhaps we can prove that given a point $P$ different from $b_0$ there exist a point $Q$ in $H^{\infty}$ such that $P,Q$ and $b_1$ are aligned and using it?

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    $\begingroup$ There are elements of $S^\infty\setminus\{b_1\}$ which have non-zero first coordinate, e.g. $(\frac{1}{2},\frac{1}{2},0,0,\dots)$. $\endgroup$ – Alex Provost Apr 4 '17 at 11:14
  • $\begingroup$ @AlexProvost arf, rigth. Do you have an idea to get the desirated homeomorphism? $\endgroup$ – JeSuis Apr 4 '17 at 12:02
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    $\begingroup$ Do you know about stereographic projection? $\endgroup$ – Andrew D. Hwang Apr 4 '17 at 13:28
  • $\begingroup$ @AndrewD.Hwang yeah a little bit. $\endgroup$ – JeSuis Apr 4 '17 at 14:51
  • $\begingroup$ First of all note that $\mathbb{R}^{\infty}\simeq H^{\infty}$. Your $f$ is the desired homeomorphism. Now you can try to approach this from the other point of view. First show that there exists a homeomorphism $f:\mathbb{S}^n\backslash\{p\}\to\mathbb{R}^n$ (a sterographic projection). Then in your construction of $\mathbb{S}^{\infty}$ remove the common point at each step and note that the family $\{f_n\}$ induces a continous mapping $f:\mathbb{S}^{\infty}\backslash\{p\}\to\mathbb{R}^{\infty}$. $\endgroup$ – freakish Apr 4 '17 at 15:08
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Ok, so lets start with a more "handy" definition of $\mathbb{R}^{\infty}$:

$$\mathbb{R}^{\infty}=\big\{(x_n)_{n=1}^{\infty}\ \big|\ x_i\in\mathbb{R},\ \exists_{N}\forall_{j>N}\ x_i=0\big\}$$

So it's a set of all real sequences that are constantly equal to $0$ at some point.

Now we can look at $\mathbb{R}^n$ as a subset of $\mathbb{R}^{\infty}$ defined by

$$\mathbb{R}^{n}=\big\{(x_n)_{n=1}^{\infty}\in\mathbb{R}^{\infty}\ \big|\ \forall_{j>n}\ x_j=0\}$$

So unlike $\mathbb{R}^{\infty}$ in $\mathbb{R}^n$ we fix the index (at $n$) starting from which a sequence has to be constantly equal to $0$.

We define topology on $\mathbb{R}^{\infty}$ by setting: $U$ is open iff $U\cap\mathbb{R}^n$ is open in $\mathbb{R}^n$ for all $n$.

All of this is consistent with your approach but formally allows us to write:

$$\mathbb{R}^1\subset \mathbb{R}^2\subset \mathbb{R}^3\subset\cdots$$ $$\mathbb{R}^\infty=\bigcup_{n=1}^{\infty}\mathbb{R}^n$$


If we write $\mathbb{S}^n$ for a unit sphere in $\mathbb{R}^n$ then it is easy to see that

$$\mathbb{S}^1\subset \mathbb{S}^2\subset \mathbb{S}^3\subset\cdots$$

and we can define

$$\mathbb{S}^{\infty}=\bigcup_{n=1}^{\infty}\mathbb{S}^n$$

which is consistent with your definition of $\mathbb{S}^{\infty}$.

Now pick a point $v\in\mathbb{S}^{\infty}$ (note that it doesn't matter if it is a part of some basis or not). Obviously there exists $N\in\mathbb{N}$ such that $v\in\mathbb{S}^{n}$ for all $n>N$. In particular

$$\mathbb{S}^{\infty}\backslash\{v\}=\bigcup_{n=N+1}^{\infty}\mathbb{S}^n\backslash\{v\}$$

Note that the indexing on the bottom starts from $N+1$. I will implicitely assume that $n>N$ from now on.

We are almost ready. Now take $\varphi_n:\mathbb{S}^n\backslash\{v\}\to\mathbb{R}^n$ to be the $n$-dimensional sterographic projection . It is well known that $\varphi_n$ are homeomorphisms and are compatible with each other in the following sense: if $n<m$ and $u\in\mathbb{S}^n\backslash\{v\}$ then $\varphi_n(u)=\varphi_m(u)$ (note that $\mathbb{R}^n\subset\mathbb{R}^m$). Actually $\varphi_n$ can be chosen arbitrarly as long as they are homeomorphisms that satisfy the compatibility property.

The compatibility property makes the following function well defined:

$$\Phi:\mathbb{S}^{\infty}\backslash\{v\}\to\mathbb{R}^{\infty}$$ $$\Phi(u)=\varphi_n(u)$$

where $n$ is such that $u\in\mathbb{S}^n$. Obviously the definition does not depend on $n$. Furthermore the continuity of $\Phi$ follows from the definition of inductive topology. Finally $\Phi^{-1}$ is given by

$$\Phi^{-1}(u)=\varphi_n^{-1}(u)$$

This function is again well defined and continous by the same argument (note that $\varphi_n^{-1}$ are also compatible on $\mathbb{R}^n$). $\Box$

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  • $\begingroup$ Are you sure the inductive topology on $\mathbb{R}^{\infty}$ is equivalent to the one given by your metric? Considering the sequence $x_n=(0,\cdots,0,1/n,0,\cdots)$, it seems that it is not convergent in the inductive topology since $\bigcup_{n} ((\prod_{i=1}^n (-1/2n,1/2n))\times \prod_{i=n+1}^{\infty}\{0\})$ seems to be a neighbourhood of $0$ which the sequence never enters, whereas the sequence $x_n$ clearly converges to $0$ on the metric you give. $\endgroup$ – Aloizio Macedo Apr 4 '17 at 18:12
  • $\begingroup$ @AloizioMacedo Indeed, defining metric was a bit too much. I've removed that part. I'm not 100% sure if continuity of $\Phi$ and $\Phi^{-1}$ works well with the induced topology but I guess it should be easy to check. $\endgroup$ – freakish Apr 4 '17 at 21:11
  • $\begingroup$ @freakish Is this topology the continuity it suffices to check the continuity of each restriction $\mathcal{S}^n\setminus\{v\}.$ How do you compute the inverse? $\endgroup$ – Alex Apr 5 '17 at 11:52
  • $\begingroup$ @Alex Yes, you are right, continuity is trivial. The inverse is given by $\Phi^{-1}(u)=\varphi_n^{-1}(u)$. This is still well defined and continous by the same argument. $\endgroup$ – freakish Apr 5 '17 at 12:38

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