# Intuitive functions for topological homeomorphims.

The basic idea behind homeomorphisms in topology is that we can continuously morph one set into another so for example :

Given $$(\Bbb R,T_{down})$$, $$(\Bbb R,T_{up})$$ where $$T_{up}=\{(a,\infty)|a\in \Bbb R\}$$,$$T_{down}=\{(-\infty,b)|b\in \Bbb R\}$$ a natural homeomorphism would be $$f(x)=-x$$, because intuitively we can just think of this as reversing the direction of a ray .

Given $$((0,1), T_{standard}),((0,\infty), T_{standard}),$$ a natural homeomorphism would be $$f(x)=\tfrac{1}{x}$$ as we can intuitively think of this as enlarging elements from the first set and shrinking elements of the second.

Given $$(\Bbb N,T_N) (\Bbb Z, T_Z)$$ where $$T_N,T_Z$$ are the subspace topologies induced from $$(\Bbb R, T_{standard})$$ , then a natural homeomorphism would be

\begin{align} \quad f(n) = \left\{\begin{matrix} -\frac{n+1}{2} & \mathrm{if} \: n \: \mathrm{if \: odd} \\ \frac{n}{2} & \mathrm{if} \: n \: \mathrm{is \: even} \end{matrix}\right. \end{align}

As we can intuitively think of sending half the number(even) to positives and half (odd) to negatives. while its inverse does the opposite(obviously).

My question is:

Could anyone provide some other natural and intuitive examples of functions (of an intermediate to beginner level ) we could use to morph sets into one another, with the idea being that the operation of the function would be somewhat obvious just from looking at in what ways the sets are different/the same.

• The intuition "continuously morphing one space into another" is more often used with homotopy. Homeomorphism is more like "these two spaces are actually the same". Another nice example is a pointed $n$-dimensional sphere being homeomorphic to $\mathbb{R}^n$ (stereographic projection). – freakish Jul 26 at 14:50
• In your $1/x$ example, you probably meant $(1,\infty)$ rather than $(0,\infty)$. – Andreas Blass Jul 29 at 18:59

The most common examples are:

• Any two bounded closed intervals $$[a,b]$$, $$[c,d]$$ of $$\mathbb{R}$$ are homeomorphic.

We can take the usual linear homeomorphism $$f\colon[a,b]\to[c,d]$$, $$f(x)=c+(x-a)\frac{d-c}{b-a}$$. Or we can take a different (quadratic) one as well, say $$g(x)=c+(x-a)^2\frac{d-c}{(b-a)^2}$$.

• Any two open intervals (even infinite) of $$\mathbb{R}$$ are homeomorphic.

You can use the same formula as above for bounded intervals.

For two unbounded intervals "in the same direction" e.g. $$(a,\infty)$$ and $$(b,\infty)$$ just take $$x\mapsto x-a+b$$. Similarly for $$(-\infty,a)$$ and $$(-\infty,b)$$.

For two unbounded intervals "in different directions", e.g. $$(-\infty,a)$$ and $$(b,\infty)$$, take $$x\mapsto-x+a+b$$.

For two intervals of the form interval $$(a,b)$$ and $$(c,\infty)$$ we can use $$x\mapsto\frac{b-a}{x-a}+c$$.

• The one-point compactification $$(a,b)_\ast$$ of any open interval $$(a,b)$$ is homeomorphic the circle $$\mathbb{S}^1$$.

Any open interval is homeomorphic to $$(-1,1)$$. The map $$x\in(-1,1)\mapsto(\cos(2\pi x),\sin(2\pi x)),\qquad\ast\mapsto(1,0)$$ (where "$$\ast$$" is the point at infinity) is a homeomorphism from $$(-1,1)_\infty$$ to $$\mathbb{S}^1$$.

• $$X\times Y$$ is homeomorphic to $$Y\times X$$.

Just take $$(x,y)\mapsto(y,x)$$.

• More generally, if $$f\colon I\to J$$ is a bijection between sets, $$\left\{X_i:i\in I\right\}$$ and $$\left\{Y_j:j\in J\right\}$$ are collections of sets such that, for all $$i\in I$$, $$X_i$$ is homeomorphic to $$Y_{f(i)}$$, then $$\prod_{i\in I}X_i$$ is homeomorphic to $$\prod_{j\in J}Y_j$$.

Take homeomorphisms $$\phi_i\colon X_i\to Y_{f(i)}$$. Then $$(y_j)_j\mapsto (\phi_i^{-1}(y_{f(i)}))_i$$ is a homeomorphism from $$\prod_J Y_j$$ to $$\prod_I X_i$$.

The homeomorphisms in the examples above have simple formulas, but sometimes some "geometric" intuition is more useful in describing some more interesting homeomorphisms (although we can still define those formally).

• The Cantor set $$\left\{0,1\right\}^\mathbb{N}$$ is homeomorphic to the ternary Cantor set of $$\mathbb{R}$$.

The usual homeomorphism is $$(x_i)_{i=1}^\infty\mapsto \sum_{i=1}^\infty 2x_i3^{-i}.$$ Intuitively, we are constructing the point associated to a sequence $$(x_i)_i$$ as follows: In the first step of the construction of the ternary Cantor set, we break the interval $$[0,1]$$ in subintervals $$[0,1/3]$$ and $$[2/3,1]$$. If $$x_1=0$$, let $$I_1$$ be the subinterval to the left. If $$x_1=1$$, let $$I_1$$ be the subinterval to the right.

Proceed in this manner inductively: If $$x_{n+1}=0$$, let $$I_{n+1}$$ be the "left third subinterval" on the left of $$I_n$$, or to the right if $$x_{n+1}=1$$. The intersection $$\bigcap_nI_n$$ contains only the point corresponding to $$(x_n)_{n=1}^\infty$$.

• [Urysohn] Any two metrizable, compact, zero-dimensional (i.e., with a basis of clopen sets) topological spaces $$X$$ and $$Y$$ without isolated points are homeomorphic.

The homeomorphism may be realized as follows:

• First fix metrics inducing the topologies of $$X$$ and $$Y$$. Break down $$X$$ and $$Y$$ into finite numbers of disjoitn clopen sets of diameter $$\leq 1$$, which is possible since they are compact and zero-dimensional. Since they do not have isolated points, we can assume they were broken down into the same number of disjoint clopen sets, say $$X_{1},\ldots,X_{N_1}$$ and $$Y_{1},\ldots,Y_{N_1}$$.
• We repeat this process with each element of each partition and iterate. Formally, given a natural number $$k$$, suppose we have numbers $$N_1,\ldots,N_k$$ and collections of sets $$X_{n_1,\ldots,n_p}$$ and $$Y_{n_1,\ldots,n_p}$$, where $$p\leq k$$ and $$1\leq n_i\leq N_i$$, such that the diamenter of $$X_{n_1,\ldots,n_p}$$ is $$\leq 1/p$$, and, if $$p\leq k-1$$, then $$\left\{X_{n_1,\ldots,n_p,1},X_{n_1,\ldots,n_p,N_{p+1}}\right\}$$ is a partition of $$X_{n_1,\ldots,n_p}$$, and similarly for the $$Y_{n_1,\ldots,n_p}$$. The same arguments as in the previous paragraph above apply, so given a sequence $$\overline{n}=(n_1,\ldots,n_k)$$, $$1\leq n_i\leq N_i$$, we can partition each of the sets $$X_{\overline{n}}$$ and $$Y_{\overline{n}}$$ into some given numberclopen sets, say $$N_{k+1}$$, which will depend only on $$k$$. We just enumerate these partition and have collection $$X_{\overline{n},1},\ldots,X_{\overline{n},N_{k+1}}$$ and $$Y_{\overline{n},1},\ldots,Y_{\overline{n},N_{k+1}}$$ of clopen sets with the same properties as above.
• Then the map $$\prod_{k=1}^\infty\left\{1,\ldots,N_k\right\}\to X$$, which takes $$(x_k)_k$$ to the unique point of $$\bigcap_kX_{x_1,x_2,\ldots,x_k}$$ is a homeomorphism, so $$\prod_{k=1}^\infty\left\{1,\ldots,n_k\right\}$$ is homeomorphic to $$X$$. Similarly, $$\prod_{k=1}^\infty\left\{1,\ldots,N_k\right\}$$ is homeomorphic to $$Y$$, so $$X$$ is homeomorphic to $$Y$$ by transitivity.
• Also, if $S$ is a subspace of $\Bbb R$ and $f:\Bbb R \to S$ is a continuous bijection then $f$ is a homeomorphism. E.g. $S=(0,\infty)$ and $f(x)=e^x$.... E.g. $S=(-1,1)$ and $f(x)=x/(1+|x|).$ But in general a continuous bijection from one space to another need not be a homeomorphism. – DanielWainfleet Jul 29 at 23:33