Imbeddings and Dimension Theory

Munkres Topology

We begin with some basic properties of topological dimension.

Theorem 1) Let $X$ be a finite dimensional space. If $Y$ is a closed subspace of $X$, $Y$ is also finite dimensional and $\dim Y\le \dim X$.

Proof) Let $m=\dim X$, and $\mathcal{A}$ be an open cover of $Y$ in the subspace topology of $Y$. For each $A\in \mathcal{A}$, choose an open set $A^{\prime }\subset X$ that is open in $X$ and such that $A=A^{\prime }\cap Y$. The sets $A^{\prime }$ and $X-Y$ form an open cover of $X$; as $X$ has dimension $m$, there is a open refinement $\mathcal{B}$ such that $\mathcal{B}$ covers $X$ with order at most $m+1$. The corresponding collection $\mathcal{C}=\{B\cap Y|B\in \mathcal{B}\}$ is the desired refinement of $\mathcal{A}$ of order at most $m+1$ in $Y$.

Theorem 2) Let $X=Y\cup Z$, where $Y$ and $Z$ are closed subspaces of $X$ with finite topological dimension. Then, $\dim X=\max \{\dim Y,\dim Z\}$.

Proof) Let $m=\max \{\dim Y,\dim Z\}$. By Theorem 1, $\dim X\ge m$; it suffices to show that $\dim X$ is at most $m$. Let $\mathcal{A}$ be an arbitrary open cover of $X$; our goal is to find an open refinement of $\mathcal{A}$ that is a cover of order at most $m+1$.

Consider the collection ${\mathcal{A}}_Y=\{Y\cap A|A\in \mathcal{A}\}$ in $Y$. As $Y$ has dimension at most $m$, there is a refinement ${\mathcal{A}}_Y^{\prime }$ of ${\mathcal{A}}_Y$ in $Y$ that is an open cover of order at most $m+1$. For every $A^{\prime }\in {\mathcal{A}}_Y^{\prime }$, choose an open set $U(A^{\prime })$ of $X$ such that $U(A^{\prime })\cap Y=A^{\prime }$ and an element $V(A^{\prime })\in \mathcal{A}$ such that $A^{\prime }\subset V(A^{\prime })$. The collection of open sets $U(A^{\prime })\cap V(A^{\prime })$ for all $A^{\prime }\in {\mathcal{A}}_Y^{\prime }$ adjoined with the collection $A-Y$ for all $A\in \mathcal{A}$ is an open cover of $X$ with order at most $m+1$ on $Y$. Call this collection $\mathcal{B}$.

Analogously, we can construct a refinement $\mathcal{C}$ of $\mathcal{B}$ such that $\mathcal{C}$ is an open cover of $X$ with order at most $m+1$ on $Z$. Define a map $f:\mathcal{C}\to \mathcal{B}$ by arbitrarily associating each $C\in \mathcal{C}$ to an element $B\in \mathcal{B}$ such that $C\subset B$, and let $D(B)$ be the union of open sets $C\in \mathcal{C}$ such that $f(C)=B$. The collection $\mathcal{D}=\{D(B)|B\in \mathcal{B}\}$ is our desired refinement. First, as $D(B)\subset B$ and $\mathcal{B}$ is a refinement of $\mathcal{A}$, implying $\mathcal{D}$ is a refinement of $\mathcal{A}$. Furthermore, $\mathcal{D}$ is clearly an open cover as $\mathcal{C}$ is an open cover. Finally, as the collection $\mathcal{B}$ has order at most $m+1$ on $Y$, $\mathcal{D}$ has order at most $m+1$ on $Y$, and as $\mathcal{C}$ has order at most $m+1$ on $Z$, $\mathcal{D}$ has order at most $m+1$ on $Z$. As $X=Y\cup Z$, $\mathcal{D}$ has order at most $m+1$ on $X$, and we are done.

Theorem 3) Every compact subspace of ${\mathbb{R}}^N$ has topological dimension at most $N$.

Proof) Define $\mathcal{J}$ to be the collection of open intervals $\{(n,n+1)|n\in \mathbb{Z}\}$ in $\mathbb{R}$ and define $\mathcal{K}$ to be the collection of one-point sets $\{\{n\}|n\in \mathbb{Z}\}$. An $M$-cube (where $M\le N$) is a set of the form $A_1\times A_2\times \dots \times A_N$ where $M$ of the sets $A_i$ are from $\mathcal{J}$ and the remaining $N-M$ are from $\mathcal{K}$. Let ${\mathcal{C}}_M$ be the collection of all $M$-cubes ($0\le M\le N$). We argue the $M$-cubes in ${\mathcal{C}}_M$ can be expanded into open sets such that they remain pairwise disjoint. Take any $M$-cube $U$ and $x\in U$; then, precisely $N-M$ coordinates of $x$ are integral while the remaining $M$ are nonintegral. Choose a radius $ϵ>0$ such that these $M$ coordinates remain nonintegral in $B(x,ϵ)$; this neighborhood $B(x,ϵ)$ cannot intersect any other $M$-cubes. Let $V_x=B(x,ϵ/2)$; expanding the $M$-cube $U$ to ${\bigcup }_{u\in U}{V}_u$ creates a new collection ${\mathcal{D}}_M$ of disjoint expanded $M$-cubes. As ${\bigcup }_{0\le M\le N}{\mathcal{C}}_M$ forms an open cover of $X$, $\mathcal{D}={\bigcup }_{0\le M\le N}{\mathcal{D}}_M$ is also an open cover such that the constituent sets of ${\mathcal{D}}_M$ are pairwise disjoint for every $M$.

Let $C\subset {\mathbb{R}}^N$ be compact and $\mathcal{A}$ be an open covering of $C$. By compactness properties, $\mathcal{A}$ permits a positive Lebesgue number $\delta >0$ such that any subset of diameter less than $\delta$ is contained in some member of the cover. Let $\mathcal{D}$ be the open cover of ${\mathbb{R}}^N$ obtained above. Let $f$ be the homeomorphism ${\mathbb{R}}^N\to {\mathbb{R}}^N$ defined as $x\mapsto x\delta /2\sqrt{N}$. Taking $f(\mathcal{D})$ then taking intersections with $C$ gives the desired refinement of $\mathcal{A}$ of order at most $N+1$ that is an open cover of $C$.

Before we move into two large theorems, we briefly explain some analytical geometry concepts central to our discussion. A set $\{x_0,x_1,\dots ,x_k\}$ of points in ${\mathbb{R}}^N$ is said to be affinely independent if the equations $\sum a_i{x}_i=0$ and $\sum a_i=0$ has the unique solution $a_i=0$ for all $0\le i\le k$. Note this is equivalent to $\{x_1-x_0,x_2-x_0,\dots ,x_k-x_0\}$ being linearly independent. If $\{x_0,x_1,\dots ,x_k\}$ is affinely independent, we call the set

$$P=\left\{\sum _{i=0}^k{t}_i{x}_i|\sum _{i=0}^k{t}_i=1\right\}$$

the $k$-plane determined by these points. $P$ is homeomorphic to the $k$-dimensional subspace spanned by $x_i-x_0$ for $1\le i\le k$, and thus if $k<N$, $P$ has an empty interior. Furthermore, it is trivial that if $x\notin P$, $\{x_0,x_1,\dots ,x_k,x\}$ is also affinely independent. We say a set $A\subset {\mathbb{R}}^N$ is in general position if every subset of $A$ containing $N+1$ or fewer points is affinely independent. If $N=3$, a set of points is affinely independent if no four points are coplanar and no three points are collinear. A key properties of sets in general position is as follows:

Theorem 4) Given a finite set $\{x_1,x_2,\dots ,x_n\}$ of points in ${\mathbb{R}}^N$ and given some $\delta >0$, there exists a set $\{y_1,y_2,\dots ,y_n\}$ of points in general position in ${\mathbb{R}}^N$ such that $|x_i-y_i|<\delta$ for all $i$.

Proof) We proceed with induction. Take $y_1=x_1$. Assume we have found points $\{y_1,y_2,\dots ,y_m\}$ for some $1\le m\le n-1$ such that they are in general position and $|x_i-y_i|<\delta$. Take the union of all the planes generated by subsets of size $N$ or less of $\{y_1,y_2,\dots ,y_m\}$; as ${\mathbb{R}}^N$ is a Baire space, this union has an empty interior and thus there is some $y_{m+1}\in B(x_{m+1},\delta )$ such that $\{y_1,y_2,\dots ,y_{m+1}\}$ is in general position.

Theorem 5) $X$ is $\sigma$-compact if there is a countable collection of compact subspaces of $X$ whose interiors cover $X$. If $X$ is a $\sigma$-compact Hausdorff space such that every compact subspace of $X$ has topological dimension at most $m$, then $X$ has dimension at most $m$.

Proof) Let $\mathcal{A}$ be an open cover of $X$. It suffices to find an open cover refining $\mathcal{A}$ that has order at most $m+1$. We begin by constructing a sequence of compact subsets $X_i$ such that $X=\bigcup X_i$, $X_0=\varnothing$, and $X_i\subset \text{Int}(X_{i+1})$. We can do so inductively, given $X_0=\varnothing$ to start; let $\{C_i{\}}_{i\ge 1}$ be a countable collection of compact subspaces of $X$ whose interiors cover $X$. Given $X_n$, the interiors of $C_i$ form an open cover of $X_n$ and thus a finite subcollection $\text{Int}(C_{i_1}),\text{Int}(C_{i_2}),\dots ,\text{Int}(C_{i_k})$ covers $X_n$. Then, take $X_{n+1}={\bigcup }_{j=1}^k{C}_{i_j}\cup C_{n+1}$; such a sequence clearly satisfies our desired properties.

Next, we take the intersections of $\mathcal{A}$ with open subsets of the form $\text{Int}(X_{n+2})-X_n$ for $n\ge 0$ to obtain an open cover ${\mathcal{B}}_0$ of $X$ that is a refinement of $\mathcal{A}$ with the additional condition that if $B\in {\mathcal{B}}_0$ and $B\cap X_n\ne \varnothing$, then $B\subset X_{n+1}$. We sequentially construct refinements of $\mathcal{A}$ as follows: suppose $n\ge 0$ and ${\mathcal{B}}_n$ is an open cover of $X$ refining ${\mathcal{B}}_0$ such that it has order at most $m+1$ on points of $X_n$. Let $\mathcal{C}$ be a refinement of ${\mathcal{B}}_n$ such that $\mathcal{C}$ is an open cover with order at most $m+1$ on points of $X_{n+1}$. Define an arbitrary map $f:\mathcal{C}\to {\mathcal{B}}_n$ such that $C\subset f(C)$ for any $C\in \mathcal{C}$, and let $D(B)$ be the union of $C\in \mathcal{C}$ such that $f(C)=B$ for any $B\in {\mathcal{B}}_n$. Let ${\mathcal{B}}_{n+1}$ consist of the sets $B\in {\mathcal{B}}_n$ for which $B\cap X_{n-1}\ne \varnothing$ and $D(B)$ for sets $B\in {\mathcal{B}}_n$ such that $B\cap X_{n-1}=\varnothing$. The collection ${\mathcal{B}}_{n+1}$, then, is clearly an open cover of $X$ that is a refinement of ${\mathcal{B}}_n$ and of order at most $m+1$ on points in $X_{n+1}$. If we let $\mathcal{B}$ be the collection of open sets $B$ such that there exists an $N$ such that $B\in {\mathcal{B}}_n$ for all $n\ge N$, $\mathcal{B}$ is the desired refinement of $\mathcal{A}$ that is an open cover with order at most $m+1$ on the entire space $X$. We are done.

Theorem 5-1) Every $m$-manifold $X$ has topological dimension at most $m$.

Proof) For every $x$ there is a neighborhood $U$ that is homeomorphic to an open subset of ${\mathbb{R}}^m$, and by normality there is a neighborhood $V$ such that $x\in V$ and $\overline{V}\subset U$. Let $f$ be this homeomorphism; take some open ball $B(f(x),ϵ)$ such that $B(f(x),ϵ)\subset f(U)$, and let $W=f^{-1}(B(f(x),ϵ/2))$. Then, $U_x=W\cap V$ is a neighborhood of $x$ such that $\overline{U_x}$ is homeomorphic to a compact subset of ${\mathbb{R}}^m$. In this way, we obtain a collection of compact subspaces of $X$ whose interiors cover $X$. As $X$ is second-countable, it is Lindelof, and there’s a countable subcollection whose interiors cover $X$. Thus, $X$ is $\sigma$-compact.

It suffices to show every compact subset of $X$ has topological dimension at most $m$. By our discussion above, there is a countable collection of compact subspace of $X$, which we denote $\{C_n\}$, whose interiors cover $X$. This collection has the additional property that each $C_n$ is homeomorphic to a compact subset of ${\mathbb{R}}^m$, and thus has topological dimension at most $m$ by Theorem 3. For any compact subset $C$ of $X$, there is a finite subcollection $C_{n_1},C_{n_2},\dots ,C_{n_k}$ of whose interiors cover $C$. Then, $C={\bigcup }_{i=1}^k(C\cap C_{n_i})$, where $C\cap C_{n_i}$ is compact and thus closed in $C$. By Theorem 2, $C$ is at most dimension $m$.

Theorem 5-2) Every closed subspace of ${\mathbb{R}}^N$ has topological dimension at most $N$.

Proof) This is trivial by Theorem 3.

Theorem 6) Let $X$ be a locally compact Hausdorff space with a countable basis, such that every compact subspace of $X$ has topological dimension at most $m$. Then $X$ is homeomorphic to a closed subspace of ${\mathbb{R}}^{2m+1}$.

Proof) We first formulate a more tractable goal in place of finding an explicit homeomorphism. If $f:X\to {\mathbb{R}}^N$ is a continuous map, we say $f(x)\to \infty$ as $x\to \infty$ if given $n$, there is a compact subspace $C$ of $X$ such that $\Vert f(x)\Vert >n$ for $x\in X-C$. Let $\overline{\rho }$ be the uniform metric in $\mathcal{C}(X,{\mathbb{R}}^N)$. Then, if $f,g\in \mathcal{C}(X,{\mathbb{R}}^N)$, $\overline{\rho }(f,g)<1$, and $f(x)\to \infty$ as $x\to \infty$, clearly $g$ must also satisfy $g(x)\to \infty$ as $x\to \infty$. We claim it suffices to find a $f\in \mathcal{C}(X,{\mathbb{R}}^N)$ that is injective and such that $f(x)\to \infty$ as $x\to \infty$ (with $N=2m+1$, in our case). This is because, if we take the one-point “compactifications” (I use quotes here because if $X$ is compact it isn’t technically a compactification) $X^{\prime }=X\cup \{{\infty }_X\}$ and ${\mathbb{R}}^{N^{\prime }}={\mathbb{R}}^N\cup \{{\infty }_{{\mathbb{R}}^N}\}$ and extend $f$ such that $f({\infty }_X)={\infty }_{{\mathbb{R}}^N}$, $f$ remains continuous. To elaborate, if $U$ is an open subset in ${\mathbb{R}}^{N^{\prime }}$ that doesn’t contain ${\infty }_{{\mathbb{R}}^N}$ then $f^{-1}(U)$ is clearly open in $X^{\prime }$, whereas if $U$ contains ${\infty }_{{\mathbb{R}}^N}$, ${\mathbb{R}}^{N^{\prime }}-U$ is closed and bounded in ${\mathbb{R}}^N$. Then, $f^{-1}({\mathbb{R}}^{N^{\prime }}-U)$ is a closed set contained in some compact set of $X$, and thus is compact in $X$. This implies $f^{-1}(U)=f^{-1}({\mathbb{R}}^{N^{\prime }}-U{)}^C$ is open in $X^{\prime }$, completing our argument. Continuing on, if $f$ is additionally injective, the extension of $f$ is a homeomorphism from $X^{\prime }$ to $f(X)\cup \{{\infty }_{{\mathbb{R}}^N}\}$. Taking the appropriate restrictions to $X$ and $f(X)$ gives us that $f$ is a homeomorphism of $X$ into some closed subspace of ${\mathbb{R}}^N$. Thus, it suffices to show such an $f$ exists.

Note that as $X$ is Hausdorff and second-countable, any compact subset $C\subset X$ is metrizable with some metric $d$. With this space and metric, let $U_{ϵ}(C)=\{f\in \mathcal{C}(X,{\mathbb{R}}^{2m+1})|\Delta f{|}_C<ϵ\}$, where, for any continuous map $g:Y\to {\mathbb{R}}^N$, $\Delta g=\sup \{\text{diam}{f}^{-1}(\{z\})|z\in g(Y)\}$. We claim that $U_{ϵ}(C)$ is dense and open in $\mathcal{C}(X,{\mathbb{R}}^{2m+1})$. Openness is quite straightforward; take some $f\in U_{ϵ}(C)$ and take some $a\in \mathbb{R}$ such that $\Delta f{|}_C<a<ϵ$. Let $A\subset C\times C$ be such that $(x,y)\in A⟺d(x,y)\ge a$. $A$ is then compact, and thus there is some $\delta >0$ such that, if $\rho$ is the square metric in ${\mathbb{R}}^{2m+1}$, $\rho (x,y)>\delta$ whenever $(x,y)\in A$. Then, the neighborhood $B(f,\delta /2)$ under the uniform metric $\overline{\rho }$ in $\mathcal{C}(X,{\mathbb{R}}^{2m+1})$ is contained in $U_{ϵ}(C)$ as any $g\in B(f,\delta /2)$ satisfies $\Delta g{|}_C\le a<ϵ$.

Now we have to show denseness, which is of an analytic geometry flavor. Set an arbitrary constant $\delta >0$ and $f\in \mathcal{C}(X,{\mathbb{R}}^{2m+1})$; it suffices to show there is some $g$ such that $\overline{\rho }(f,g)<\delta$ and such that $g\in U_{ϵ}(C)$; we do so via explicit construction with partitions of unity. In the subspace topology of $C$, we can find a finite open covering $\{U_i{\}}_{1\le i\le k}$ such that $\text{diam }(U_i)<ϵ$, $\text{diam }(f(U_i))<\delta /2$, and such that $U_i$ has order at most $m+1$. Construct a partition of unity $\{{\varphi }_i{\}}_{1\le i\le k}$ dominated by $\{U_i\}$. Take arbitrary points $x_i\in U_i$; by Theorem 4, we can find points $z_i\in {\mathbb{R}}^{2m+1}$ corresponding to $U_i$ such that $\{z_i{\}}_{1\le i\le k}$ is in general position in ${\mathbb{R}}^{2m+1}$ and $\rho (z_i,f(x_i))<\delta /2$. Take the function $g:C\to {\mathbb{R}}^{2m+1}$ defined as follows:

$$g(x)=\sum _{i=1}^k{\varphi }_i(x)z_i$$

We claim this leads to our desired function in $U_{ϵ}(C)$. First, $\overline{\rho }(f,g)<\delta$ as for any $x\in X$, $g(x)$ is the weighted average of finitely many points $z_i$ that deviate at most $\delta$ from $f(x)$; thus, $\rho (g(x),f(x))<\delta$ for all $x\in X$. Furthermore, assume there exists $x,y\in X$ such that $g(x)=g(y)$, or $g(x)-g(y)=0$. As $U_i$ has order at most $m+1$, there can be at most $2m+2$ distinct $z_i$s with nonzero coefficients in this sum. Furthermore, the sum of all coefficients is $\sum {\varphi }_i(x)-{\varphi }_i(y)=0$. However, $\{z_i\}$ is in general position in ${\mathbb{R}}^{2m+1}$, meaning the only plausible solution is for all the coefficients of each $z_i$ to be zero; that is, ${\varphi }_i(x)={\varphi }_i(y)$ for all $1\le i\le k$. This means there must be some $U_i$ such that $x$ and $y$ are both in $U_i$, implying $d(x,y)<ϵ$. Therefore, $\Delta g<ϵ$.

We now have to extend $g$ to be defined on $X$. Let $h=g-f$; by our previous discussion, $\overline{\rho }(h)<\delta$. By Tietze extension and properties of product topologies, there is some extension $h^{\prime }$ of $h$ defined on all of $X$ such that $\overline{\rho }(h^{\prime })<\delta$; taking $g^{\prime }=f+h^{\prime }$ gives us our desired function in $U_{ϵ}(C)$.

The function space $\mathcal{C}(X,{\mathbb{R}}^{2m+1})$ under the uniform topology is metrizable and thus a Baire space. $X$ is clearly $\sigma$-compact, and following Theorem 5 we can take a countable collection of compact spaces $C_i$ such that $C_0=\varnothing$, $C_i\subset \text{Int}(C_{i+1})$, and $X=\bigcup C_i$. By Baire space properties, $U=\bigcap U_{1/n}(C_n)$ is dense in $\mathcal{C}(X,{\mathbb{R}}^{2m+1})$. Let $f_n$ be a continuous function such that $f_n(C_n)=0$ and $f_n(X-\text{Int}(C_{n+1}))=\vec{1}$; then $f={\sum }_{i=0}^{\infty }{f}_i$ is an element of $\mathcal{C}(X,{\mathbb{R}}^{2m+1})$ such that $f\to \infty$ as $x\to \infty$. By the denseness of $U$, there is some $g\in U$ such that $\overline{\rho }(g,f)<1$. Then, $g$ is an injective function such that $g(x)\to \infty$ as $x\to \infty$; we are done.