Every smooth manifold embeds in some finite-dimensional Euclidean space

If $M=\mathrm{\varnothing }$, take the empty maps; all conclusions hold. Otherwise choose a countable smooth atlas as follows. Fix a countable topological basis and form all smooth chart tuples. For each nonempty basis member contained in a chart domain, [A1] selects one such tuple. These chart domains cover $M$ because every point lies in a basis member inside some chart. Enumerate them as $(U_j,x_j)$, repeating one if the family is finite. This construction records chart witnesses rather than assuming a countable-atlas theorem with missing choices.

We first prove a Euclidean estimate for later use. Let $f:D\to {\mathbb{R}}^b$ be $C^1$, where $D\subseteq {\mathbb{R}}^a$ is open and $0<a<b$. Its domain is covered by the countable family of positive-side rational closed cubes $Q\subseteq D$, by [L4]. For a fixed such cube choose a slightly larger closed cube $Q^{\mathrm{\prime }}$ with $Q\subseteq \mathrm{int}{Q}^{\mathrm{\prime }}\subseteq Q^{\mathrm{\prime }}\subseteq D$. Compactness and continuity of the finitely many derivative entries bound the derivative norm on $Q^{\mathrm{\prime }}$ by some finite $L\ge 1$. Apply [L7] on the convex OPEN set $\mathrm{int}{Q}^{\mathrm{\prime }}$ to see that $f{\mathrm{\mid }}_Q$ is $L$-Lipschitz. If $Q$ has side $\mathrm{\ell }$, subdividing into $s^a$ cubes gives an image cover by $s^a$ closed $b$-cubes of side $2L\sqrt{a}\mathrm{\ell }\mathrm{/}s$, centred at images of the prescribed lower corners. Its total volume is $(2L\sqrt{a}\mathrm{\ell }{)}^b{s}^{a-b},$ which tends to zero. Thus $f(Q)$ has content zero.

A nonempty open subset of ${\mathbb{R}}^b$, $b\ge 1$, is not null. Indeed it contains a positive-side closed cube $Q_0$. If this cube were null, [L6] would give finite cube covers with arbitrarily small total volume. Every finite cube cover of $Q_0$ has total volume at least $\mathrm{vol}(Q_0)$: subdivide a bounding rectangle at all cover-cube and $Q_0$ endpoint coordinates. Each positive-volume cell in $Q_0$ lies in some covering cube, as is seen at its midpoint, and can be assigned to the least such index. By the finite volume additivity in [L5], the sum of volumes assigned to one covering cube is at most its volume, while all cells of $Q_0$ have total volume $\mathrm{vol}(Q_0)>0$. This contradicts content zero.

Choose $\rho \ge 0$ from [L1]. For each integer $m\ge 1$ set $K_m={\rho }^{-1}([m-1,m+2]),O_m={\rho }^{-1}((m-5\mathrm{/}4,m+9\mathrm{/}4))\mathrm{.}$ The sets $K_m$ are compact by [L1] and [L2], they cover $M$, and $K_m\subseteq O_m$. The open sets $O_m$ and $O_{m^{\mathrm{\prime }}}$ are disjoint when $m\mathrm{\ne }{m}^{\mathrm{\prime }}$ and $m\equiv m^{\mathrm{\prime }}(\mathrm{m}\mathrm{o}\mathrm{d}4)$. The entire family $(O_m)$ is locally finite: the open neighbourhood ${\rho }^{-1}((\rho (p)-1,\rho (p)+1))$ can meet $O_m$ only for integers $m$ in a bounded real interval, hence only finitely many $m$.

Taking the countable union over those rational cubes and using [L5] shows that $f(D)$ is null. If $a=0<b$, $D$ is empty or a singleton and the same conclusion follows from a degenerate zero-volume cube. In particular this estimate applies to any $C^1$ map on an open subset of ${\mathbb{R}}^{2n}$ with values in ${\mathbb{R}}^{d-1}$ when $d>2n+1$. It uses only Euclidean cube nullity, not atlas invariance of manifold null sets or an equidimensional Lipschitz theorem on the wrong domain.

For a fixed nonempty $K_m$, take all tuples $(U,x,\chi ,V)$ in which $(U,x)$ is a relatively compact smooth coordinate ball with $\bar{U}\subseteq O_m$, $\chi :M\to [0,1]$ is smooth with support in $U$, and $V\subseteq U$ is open with $\chi =1$ on $V$. Their $V$-sets cover $K_m$: at a fixed point choose a small chart and apply [L3] to the compact singleton in its domain. By [L2] there is a finite ordered list of such tuples whose $V$-sets cover $K_m$. This is a finite subcover of the set of ALL eligible data, not a choice of data at every point.

We now prove the required projection assertion locally. Let $K\subseteq M$ and let $h:M\to {\mathbb{R}}^d$ be smooth, injective on $K$, with $d{h}_p$ injective for every $p\in K$, where $d>2n+1$. Write $h=(h^{\mathrm{\prime }},h_d)$ with $h^{\mathrm{\prime }}$ taking values in ${\mathbb{R}}^{d-1}$. For a parameter $v\in {\mathbb{R}}^{d-1}$ define $L_v(z^{\mathrm{\prime }},z_d)=z^{\mathrm{\prime }}-v{z}_d$. Bad secant parameters are values of $\frac{h^{\mathrm{\prime }}(q)-h^{\mathrm{\prime }}(p)}{h_d(q)-h_d(p)}$ wherever the denominator is nonzero. In every chart pair from step 1.1 this is a smooth map on an open subset of ${\mathbb{R}}^{2n}$. By steps 2.1 and [L5], its image, united over countably many chart pairs, is null.

In a source chart with coordinates $x$, write $h_j=h\circ x_j^{-1}$. Bad tangent parameters are values of $\frac{D{h}_j^{\mathrm{\prime }}(x)w}{D{h}_{d,j}(x)w}$ on the open subset of $(x_j(U_j))\times {\mathbb{R}}^n$ where the denominator is nonzero. This is smooth because $h$ is smooth, and its domain has dimension $2n$. Steps 2.1 and [L5] make the union of these images over all chart indices null. Thus the union of bad secant and tangent parameters is null in ${\mathbb{R}}^{d-1}$ and cannot be all of it by step 1.3. Choose $v$ outside that union.

For one such finite list, define each coordinate-bump block on $U$ by $(\chi ,\chi x^1,\dots ,\chi x^n)$ and by zero outside $U$. It is smooth: on the open cover $U$ and $M\setminus \mathrm{supp}\chi$ the formulas are smooth and agree; continuity follows by the same open-cover argument. Concatenate the finitely many blocks into $B_m:M\to {\mathbb{R}}^{q_m}$. Its support is contained in a finite union of compact supports inside $O_m$. If $B_m(p)=B_m(q)$ with $p,q\in K_m$, a block whose $\chi$ equals one near $p$ also has $\chi (q)=1$, and equality of its chart coordinates gives $p=q$. At $p\in K_m$ that block's differential includes the invertible chart differential, so $d{B}_{m,p}$ is injective.

The map $L_{v}h$ is injective on $K$. For if its values at $p,q\in K$ agree and $h_d(p)\mathrm{\ne }{h}_d(q)$, the quotient in step 3.1 equals $v$, a contradiction. If $h_d(p)=h_d(q)$, the equality also gives $h^{\mathrm{\prime }}(p)=h^{\mathrm{\prime }}(q)$, hence $p=q$. Similarly, if $d(L_{v}h{)}_p(w)=0$ for $p\in K$ and a tangent vector $w$, a nonzero last component of $d{h}_p(w)$ would make $v$ a bad tangent parameter. A zero last component forces all of $d{h}_p(w)$ to be zero, so $w=0$. This proves the immersion property on $K$. Notice that $h$ was never assumed to be a global embedding.

The continuous map $B_m$ is bounded. To see this without assuming a metric on $M$, its compact support is covered by the increasing open sets $\{p:\mathrm{\parallel }{B}_m(p)\mathrm{\parallel }<r\}$, $r\ge 1$ an integer; a finite subcover gives one bound, and $B_m=0$ off the support. Choose the least integer $c_m\ge 1$ strictly bounding its norm and set $H_m=2^{-m}{c}_m^{-1}{B}_m$. Then $\mathrm{\parallel }{H}_m\mathrm{\parallel }\le 2^{-m}$, its support lies in $O_m$, and it retains both separation properties on $K_m$. If $K_m$ is empty take $H_m=0$ into $\mathbb{R}$, with the properties vacuous.

Replace $L_v$ by $T_v=(1+\mathrm{\parallel }v\mathrm{\parallel }{)}^{-1}{L}_v$. This nonzero rescaling preserves both properties and is a contraction, since $\mathrm{\parallel }{L}_v(z^{\mathrm{\prime }},z_d)\mathrm{\parallel }\le \mathrm{\parallel }{z}^{\mathrm{\prime }}\mathrm{\parallel }+\mathrm{\parallel }v\mathrm{\parallel }\mathrm{\mid }{z}_d\mathrm{\mid }\le (1+\mathrm{\parallel }v\mathrm{\parallel })\mathrm{\parallel }(z^{\mathrm{\prime }},z_d)\mathrm{\parallel }\mathrm{.}$ Repeat the proved assertion a finite number of times until the target dimension is $D=2n+1$; if the initial dimension is smaller, pad by zero coordinates. Thus every such $h$ has a linear contraction $T$ into ${\mathbb{R}}^D$ preserving injectivity and injective differential on $K$. Each finite list of admissible parameters exists by finite induction; no countably dependent selection has been used.

Use [A1] on the countable family of nonempty sets of finite covering lists from step 2.2 to obtain the maps $H_m$ simultaneously. For each resulting $H_m$, step 5.1 supplies a nonempty set of finite projection certificates; a second application of [A1] selects one certificate for each $m$. Let $T_m$ be its contraction into the fixed ${\mathbb{R}}^D$, and put $C_m=T_m{H}_m$. Then $C_m$ is smooth, supported in $O_m$, bounded by $2^{-m}$, injective on $K_m$, and has injective differential there. This explicitly accounts for the countably many variable-length finite projection constructions.

For $r=0,1,2,3$ put $G_r={\sum }_{\begin{array}{c}m\ge 1\\ m\equiv r(4)\end{array}}{C}_m,G=(G_0,G_1,G_2,G_3)\mathrm{.}$ Local finiteness from step 1.4 and [L3] make these maps smooth componentwise. In each colour at most one term is nonzero at any point, by the disjointness of the $O_m$. Thus $\mathrm{\parallel }{G}_r\mathrm{\parallel }\le 1$ and $\mathrm{\parallel }G\mathrm{\parallel }\le 2$ on all of $M$. Define $J=(G,\rho )$.

If $J(p)=J(q)$, the equality $\rho (p)=\rho (q)$ puts both points in some common $K_m$, since the intervals defining the $K_m$ cover $[0,\mathrm{\infty })$. On $O_m$ the colour $m<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>4$ component of $G$ is exactly $C_m$, all other terms of that colour vanishing there. Hence $C_m(p)=C_m(q)$ and step 6.1 gives $p=q$. At a point $p\in K_m$, the same component agrees with $C_m$ on the open neighbourhood $O_m$, so its differential is injective. Therefore $d{J}_p$ is injective. This proves that $J$ is an injective immersion.

The map $J$ is proper. For a compact subset $B$ of its Euclidean target, compactness bounds the last coordinate above by some $R$ and makes $B$ closed. Thus $J^{-1}(B)$ is a closed subset of the compact sublevel set $\{\rho \le R\}$, hence compact by [L1] and [L2].

Finally $J$ is a homeomorphism onto its image. Indeed, for a closed subset $A\subseteq M$ and a point $z\mathrm{\notin }J(A)$, take the closed Euclidean unit ball $B$ about $z$. By step 8.2 the set $A\cap J^{-1}(B)$ is compact. Its continuous image $C$ is compact: pull back any ambient open cover, take a finite subcover, and map it forward. Thus $C$ is closed by [L2]. The open neighbourhood $\mathrm{int}B\setminus C$ of $z$ misses $J(A)$, showing that $J(A)$ is closed in the ambient Euclidean space. Hence the continuous bijection $J:M\to J(M)$ is closed and has continuous inverse. Together with step 8.1, [F2] makes $J$ a smooth embedding. The bounds in step 7.1 and the exhaustion from step 1.4 give all the claimed bounded-plus-proper conclusions, including the empty case.