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To start with, let me just state this theorem:

THEOREM 1 Let $(X_i,d_i)$,$(Y_i,d_i^{\,\prime})$ be metric spaces for $i=1,\dots,n$. Let $f_i:X_i\to Y_i$ be continuous for each $i$ as functions from the metric spaces $X_i$ to $Y_i$. Define the metric spaces $(X,d)$ and $(Y,d^{\,\prime})$ by setting $X=\prod_{i=1}^n X_i$ and $Y=\prod_{i=1}^n Y_i$, $d=\max{d_i}$ and $d^{\,\prime}=\max \{d_i^{ \,\prime}\}$. Then the function $f:X\to Y$ is continuous, where $f(x)=(f_1(x_1),\dots,f_n(x_n))$.

Now I am facing this seemingly simple exercise:

For each pair of points $a,b\in \Bbb R^n$ prove that there is a topological equivalence between $(\Bbb R^n,d)$ and itself defined by the inverse functions $f:\Bbb R^n\to\Bbb R^n$ and $g:\Bbb R^n\to\Bbb R^n$ such that $f(a)=b$. Hint: If $a=(a_1,\dots,a_n)$,$b=(b_1,\dots,b_n)$ define $f$ by $f(x)=(x_1+b_1-a_1,\dots,x_n+b_n-a_n)$.

... where $d$ is the $\max$ metric, that is $d(x,y)=\max\limits_{1\leq i \leq n}\{|x_i-y_i|\}$.

By the theorem it is evident $f$ is continuous. Setting $g(x)=(x_1+a_1-b_1,\dots,x_n+a_n-b_n)$ gives the desired inverse, which by the theorem is again continuous.

Could someone explain to me the motivation of this, give an insight of what this is telling us? What is the interpretation of setting $f(a)=b$? Are we setting $f(a)=b$ for each pair (in a one-one onto fashion), or just for a fixed pair $(a,b)\in \Bbb R^n\times \Bbb R^n$? Could you give another example of $f:\Bbb R^n\to \Bbb R^n \,/\,f(a)=b$?

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You are showing that given any particular $a$ and $b$, there is a homeomorphism that maps that particular $a$ to that particular $b$. It's telling you that from the metric point of view, $\mathbb{R}^n$ "looks the same" from any of its points. That is, there are no "distinguished points" in $\mathbb{R}^n$. (This is not true in, say, the vector space point of view, where the origin is "special"; it is also not the case in the metric space $[0,1]$, where $0$ and $1$ are "different" from any point in $(0,1)$, etc.) –  Arturo Magidin Jul 20 '12 at 19:43
    
@ArturoMagidin So we're focusing primarily on $a\mapsto b$, although the aforementioned "homeomorphism" is indeed from $\Bbb R^n$ onto itself. –  Pedro Tamaroff Jul 20 '12 at 19:46
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@PeterTamaroff Yes, the point is that there are homeomorphisms of $\mathbb{R}^n$ that send any chosen point to any other chosen point, so no points are "special" topologically speaking, as Arturo as said. This is different to other concepts of homomorphism, eg in various algebraic structures, identities must be sent to identities. –  Ragib Zaman Jul 20 '12 at 19:49
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@Peter: The homeomorphism is of the entire space to itself, and it has the particular property of mapping $a$ to $b$. Because $a$ and $b$ are arbitrary, it's telling you that if you are "standing" on $b$ and looking around the space, you will 'see" the same thing (via $f$ and $f^{-1}$) as I will if I'm "standing" in $a$. Again, think about $[0,1]$: given any $a,b$ in $(0,1)$ you can find a homeomorphism $[0,1]\to[0,1]$ that maps $a$ to $b$, but you cannot do it if $a=0$ and $b\notin\{0,1\}$ (because removing $a$ does not disconnect the space, but removing $b$ does). –  Arturo Magidin Jul 20 '12 at 19:49
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Essentially, what you are seeing is that the space $\mathbb R^n$ is uniform in the sense that there is nothing about a point that makes it "look" topologically different from any other point. Consider a figure eight. Any self-homeomorphism of a figure eight to itself has to send the "cross" point to itself - that point has a distinct role in the entire space that makes it "different" from the other points. –  Thomas Andrews Jul 20 '12 at 19:51
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1 Answer 1

up vote 8 down vote accepted

The theorem tells us that $\Bbb R^n$ is homogeneous: for any points $a,b\in\Bbb R^n$ there is a surjective homeomorphism $h_{a,b}:\Bbb R^n\to\Bbb R^n$ carrying $a$ to $b$, i.e., such that $h_{a,b}(a)=b$. In general $h_{a,b}\ne h_{c,d}$ if $\langle a,b\rangle\ne\langle c,d\rangle$.

Intuitively speaking, a homogeneous space is one in which all points look the same: there is no topological property that distinguishes one point from another. The circle $S^1$ is another homogeneous space: for any $a,b\in S^1$ we can take $h_{a,b}$ to be any map that rotates the circle about its centre through an angle that takes $a$ to $b$.

A less obvious example of a homogeneous space is the middle-thirds Cantor set $C$. Fix $a,b\in C$; $a$ and $b$ have ternary expansions containing no $1$s, so we can write

$$a=\sum_{n\ge 1}\frac{a_n}{3^n}\quad\text{ and }\quad b=\sum_{n\ge 1}\frac{b_n}{3^n}\;,$$

where $a_n,b_n\in\{0,2\}$ for $n\in\Bbb Z^+$. Define $h_{a,b}:C\to C$ as follows. If $c=\sum_{n\ge 1}\frac{c_n}{3^n}$ with $c_n\in\{0,2\}$ for $n\in\Bbb Z^+$, let

$$\widehat{c_n}=\begin{cases}c_n,&\text{if }a_n=b_n\\2-c_n,&\text{if }a_n\ne b_n\;,\end{cases}$$

and let $$h_{a,b}(c)=\sum_{n\ge 1}\frac{\widehat{c_n}}{3^n}\;.$$

Clearly $h_{a,b}(a)=b$, and it’s a nice little exercise to verify that $h_{a,b}$ is a surjective homeomorphism. (You might find this a little surprising, since at first glance it might appear that the endpoints of the deleted middle thirds are somehow different from the other points of $C$.)

On the other hand, the closed unit interval $[0,1]$ is not homogeneous: there is no surjective homeomorphism $h:[0,1]\to[0,1]$ taking either of the endpoints to a point in $(0,1)$. This is because the endpoints are distinguished from the other points by the following topological property: if $x\in(0,1)$, then $[0,1]\setminus\{x\}$ is not connected, but if $x\in\{0,1\}$, then $[0,1]\setminus\{x\}$ is connected. (The points of $(0,1)$ are said to be cut points of $[0,1]$.) As another exercise you might try to show that the property of being a cut point is preserved by homeomorphisms.

Even more extreme types of inhomogeneity are possible. Let $X_0$ be a closed line segment in $\Bbb R^2$ of length $1$, and let $x_0$ be its midpoint. Form $X_1$ by attaching a segment of length $1/2$ to $x_0$ that meets $X_0$ only at $x_0$. Let $x_1,x_2$, and $x_3$ be the midpoints of the ‘arms’ of $X_1$. Form $X_2$ by attaching two segments to $x_1$, three to $x_2$, and four to $x_3$ in such a way that the new ‘arms’ have length $\le1/4$, are pairwise disjoint, and do not intersect $X_1$ except at their points of attachment. Continue in this fashion: given $X_n$ with last point of attachment $x_m$ and $k$ ‘arms’, let $x_{m+1},\dots,x_{m+k}$ be the midpoints of those arms, and for $i=1,\dots,k$ attach $m+i+1$ segments of length $\le 2^{-k}$ to $x_{m+i}$ in such a way that the new ‘arms’ are pairwise disjoint and meet $X_n$ only at their respective points of attachment. The resulting set is $X_{n+1}$.

Now let $X=\bigcup_{n\in\Bbb N}X_n$, and let $D=\{x_n:n\in\Bbb N\}$. $X$ inherits its topology and metric from $\Bbb R^2$, and it’s not hard to check that $D$ is a dense subset of $X$. For each $n\in\Bbb N$, $X\setminus\{x_n\}$ has $n+2$ components. The number of connected components of $X\setminus\{x\}$ is a topological property of $x$: it’s preserved by homeomorphisms. Thus, no homeomorphism $h:X\to X$ can take $x_m$ to $x_n$ if $m\ne n$. Moreover, if $x\in X\setminus D$, then $X\setminus\{x\}$ has two components, so no homeomorphism $h:X\to X$ can take any point of $D$ to a point of $X\setminus D$. Thus, if $h:X\to X$ is a homeomorphism, $h(x_n)=x_n$ for every $n\in\Bbb N$. Now use the fact that $h$ is continuous and $D$ is dense in $X$ to show that $h(x)=x$ for every $x\in X$. In other words, the identity map is the only homeomorphism of $X$ to itself: no two distinct points of $X$ ‘look alike’ in a topological sense.

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Prof. Scott: This is awesome! –  Pedro Tamaroff Jul 21 '12 at 1:27
    
@Peter: Glad you like it! –  Brian M. Scott Jul 21 '12 at 1:30
    
@Brian: The Cantor set example is marvalous! How do you discover it ? –  Yuchen Liu Jul 21 '12 at 4:03
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@jerrysciencemath: It’s actually very natural if you know that the middle-thirds Cantor set is homeomorphic to the Cartesian product $X=\prod_n X_n$ of countably infinitely many $2$-point discrete spaces $X_n$. If $X_n=\{0,2\}$, points of $X$ are sequences $\langle x_n:n\in\Bbb Z^+\rangle$, and the map $\langle x_n:n\in\Bbb Z^+\rangle\mapsto\sum_{n\ge 1}\frac{x_n}{3^n}$ is the desired homeomorphism. Now if $D=\{n:a_n\ne b_n\}$, the map that I defined in my answer is really just flipping each $X_n$ with $n\in D$ ‘upside-down’, i.e., interchanging the $0$ and $2$. But this is still ... –  Brian M. Scott Jul 21 '12 at 4:14
    
... the product of countably infinitely many discrete $2$-point spaces, so it’s the ‘same’ space. –  Brian M. Scott Jul 21 '12 at 4:14
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