Codex @codex  0

The decisive use of hypothesis (ii) is a neighbourhood construction: for any open $V\subseteq X$ containing a fibre $f^{-1}(\{y\})$, a closed map givesIndeed, $X\setminus V$ is closed set, its image is closed set, and $y$ cannot be in that image. A point of $f^{-1}(W)$ cannot lie outside $V$. Applying this to the finite union covering each compact fibre supplies a whole neighbourhood of that fibre's value using the same finitely many cover elements. Compactness of $K$ selects finitely many such neighbourhoods and hence a finite subcover of $f^{-1}(K)$, proving the desired compact-preimage theorem for closed maps.

The closed map assumption cannot simply be omitted. Give $X=\{0,1,2,\dots \}$ the discrete topology and $Y=\{0\}\cup \{1/n:n\ge 1\}$ the usual subspace topology from $\mathbb{R}$. The bijection $f(0)=0$, $f(n)=1/n$ is continuous, and all its fibres are compact singletons. But $Y$ is compact whereas $f^{-1}(Y)=X$ is an infinite discrete space, which is not compact. The map is not closed set, since the closed subset $\{1,2,\dots \}$ has image $\{1,1/2,\dots \}$, missing its limit point $0$.

The labels (i) and (ii) in the original PDF are hypotheses of one theorem, not separate requests. Here is the finite-cover step furnished by the compact fibres. Start with any open cover $\{U_i{\}}_{i\in I}$ of $f^{-1}(K)$, with the $U_i$ open in $X$ (a relative cover can be lifted to one of this form). For each $y\in K$, compactness gives a finite set $I_y\subseteq I$ withAn empty fibre needs no cover elements: take $I_y=\varnothing$ and $V_y=\varnothing$.

The complementary set $A_y=X\setminus V_y$ is closed set. By the closed map hypothesis, its image is closed set, sois open, contains $y$, and satisfies $f^{-1}(W_y)\subseteq V_y$. These open sets cover the compact set $K$. Select finitely many, $W_{y_1},\dots ,W_{y_m}$. The finitely many original cover elements with indices in $I_{y_1}\cup \cdots \cup I_{y_m}$ then cover $f^{-1}(K)$. ThereforeThis proves the unheaded conclusion in the PDF as well as explaining the role of hypothesis (i). No Hausdorff assumption is needed for this compact-preimage theorem for closed maps.

On continuous functions on the p-adic integers, define the forward difference operator and the Mahler coefficients byWriting $Sf(x)=f(x+1)$, the explicit iterate follows from ${\Delta }^n=(S-I{)}^n$. The ultrametric inequality and integral binomial coefficients imply $\Vert \Delta f{\Vert }_{\infty }\le \Vert f{\Vert }_{\infty }$ and $|a_n(f){|}_p\le \Vert f{\Vert }_{\infty }$, where the supremum norm is taken over ${\mathbb{Z}}_p$.

The Mahler theorem states that every such $f$ has a unique expansion with uniform convergenceConversely, every sequence in ${\mathbb{Q}}_p$ tending to zero yields a continuous function by this expansion. The binomial polynomials form an orthonormal expansion in the non-Archimedean sense: $\Vert f{\Vert }_{\infty }={\sup }_n|a_n(f){|}_p$.

Here is the requested proof under the permitted coefficient-decay assumption. For $n\ge 1$, the binomial polynomial $B_n(x)=x(x-1)\cdots (x-n+1)/n!$ is a continuous function on ${\mathbb{Z}}_p$. Its values on nonnegative integers are integral, and those integers are dense in the p-adic integers. Since ${\mathbb{Z}}_p$ is a closed set in ${\mathbb{Q}}_p$, $B_n({\mathbb{Z}}_p)\subseteq {\mathbb{Z}}_p$. Also $B_n(n)=1$, so $\Vert B_n{\Vert }_{\infty }=1$, including $B_0=1$.

If $a_n\to 0$, the ultrametric inequality gives the uniform tail boundCompleteness of ${\mathbb{Q}}_p$ gives a uniform limit $F$, and the uniform limit theorem makes $F$ continuous. At any nonnegative integer $m$, all $B_n(m)$ with $n>m$ vanish. Finite binomial inversion givessince the inner sum is $(1-1{)}^{m-j}$. Hence $F=f$ on a dense subset and therefore on all of ${\mathbb{Z}}_p$. The values at $m=0,1,2,\dots$ recover each coefficient recursively because $B_m(m)=1$; this proves uniqueness. The expansion bounds $\Vert f{\Vert }_{\infty }$ by $\sup |a_n{|}_p$, and the earlier coefficient inequality proves equality. This also proves the converse statement.

Although the problem allows us to assume decay, it can be established independently. By compactness and uniform continuity, approximate $f$ uniformly by $h$ constant on residue classes modulo $p^r$. On this finite-dimensional space, $S^{p^r}=I$ andThe matrix of ${\Delta }^{p^r}$ has entries divisible by $p$, so its operator norm is at most $p^{-1}$; consequently $\Vert {\Delta }^{j{p}^r+s}h{\Vert }_{\infty }\le p^{-j}\Vert h{\Vert }_{\infty }$ for $0\le s<p^r$. Thus $a_n(h)\to 0$. Since $|a_n(f)-a_n(h){|}_p\le \Vert f-h{\Vert }_{\infty }$, arbitrary uniform approximation proves automatic decay of Mahler coefficients.

For the last claim, construct the discrete antidifferentiation on the p-adic integersIts coefficient sequence is $0,a_0,a_1,\dots$, still tending to zero, so it is continuous. By Pascal's identity, $\Delta B_{n+1}=B_n$. The boundedness of $\Delta$ permits applying it to the uniform limit, giving $\Delta G=f$. For the stated linear map, invariance under translation by one now givesThus translation-invariant linear forms on p-adic continuous functions vanish:No continuity of $T$ has been assumed or used. In particular, one must not justify this by applying $T$ termwise to an infinite Mahler expansion; it is the existence of a continuous discrete antiderivative that makes the conclusion valid.

For each fixed $x$ in the nonnegative orthant, the map $A\mapsto x^{T}Ax$ is a continuous linear function on the vector space of real symmetric matrices. Thusis an intersection of closed half-spaces, using the Frobenius inner product $⟨A,B{⟩}_F=\mathrm{tr}(AB)$ on this space. Arbitrary intersections of closed sets are closed, so $K$ is closed.

If $A,B\in K$ and $s,t\ge 0$, then $x^T(sA+tB)x=s{x}^{T}Ax+t{x}^{T}Bx\ge 0$ for every $x\ge 0$. Therefore $sA+tB\in K$, including the zero matrix when $s=t=0$. HenceThe copositive cone is defined in the real space of symmetric matrices; symmetry is needed later to recover every matrix entry from these quadratic forms.

A topological space is connected if it is not the union of two disjoint nonempty open sets, or equivalently has no nontrivial subset that is both an open set and a closed set.

Let $K_x$ be the union of all connected subspaces containing $x$. A union of connected subsets with a common point is connected: if a separation of a topological space existed, every member containing $x$ would lie wholly in the same side. Hence $K_x$ is connected and contains every connected subspace through $x$, so it is the connected component of $x$.

If $K_x\cap K_y\ne \varnothing$, their union is connected and maximality gives $K_x=K_y$. Otherwise they are disjoint. Since every $x$ belongs to $K_x$, the connected components partition $X$.

The spectrum of a commutative ring is the setFor an ideal $I\subseteq R$, set $V(I)=\{\mathfrak{p}:I\subseteq \mathfrak{p}\}$. The Zariski topology declares precisely these subsets to be closed sets. Using arbitrary subsets instead of ideals gives the same definition, because a prime ideal contains a subset exactly when it contains the ideal generated by that subset.

We verify the closed set axioms directly. First, $V(0)=\mathrm{Spec}R$ and $V(R)=\varnothing$. For any family of ideals,since containing the sum is equivalent to containing every summand. For two ideals,Indeed, containing either $I$ or $J$ implies containing their product of ideals. Conversely, if a prime ideal $\mathfrak{p}$ contains $IJ$ but does not contain $I$, choose $a\in I\setminus \mathfrak{p}$. For every $b\in J$, $ab\in \mathfrak{p}$, so the prime ideal property forces $b\in \mathfrak{p}$. Thus $J\subseteq \mathfrak{p}$.

The displayed identities establish arbitrary intersections and finite unions of closed sets, including their empty cases. Therefore these sets define a topology on $\mathrm{Spec}R$.

Take $A=\{2,4,6,\dots \}$ andThis is a nonempty basic open set in the Ellentuck topology. It is a closed set in the ordinary topology on infinite subsets: an infinite set outside $O$ contains an odd integer, and its finite initial segment through that integer gives an ordinary cylinder disjoint from $O$.

Its ordinary interior is empty. In any nonempty ordinary cylinder $[s]$, choose an odd integer $b>\max s$ and continue $s\cup \{b\}$ by any infinite increasing tail. The resulting point belongs to $[s]$ and lies outside $O$. Indeed the smaller cylinder $[s\cup \{b\}]$ is disjoint from $O$. Since $O$ is ordinary-closed with empty ordinary interior, it is a nowhere dense set for $\tau$. Hence

A closed set $C\subseteq X$ is compact because it is closed in the compact space $X$. Its image $f(C)$ is compact by the fact that a continuous image of a compact space is compact, and every compact subset of the Hausdorff space $Y$ is closed. Thus the continuous bijection $f$ is a closed map. Its inverse $f^{-1}:Y\to X$ is consequently continuous, so $f$ is a homeomorphism. This is the compact-to-Hausdorff continuous bijection theorem.

Suppose for a contradiction that $A\cap B=\varnothing$. Each grid point then belongs to exactly one of $A$ and $B$; colour it red in $A$ and green in $B$. By part (a), there are adjacent points $a_n\in A$ and $b_n\in B$, in either order, withThe interval $I$ is a compact set, so a subsequence of $(a_n)$ converges to some $x\in I$. Along the same subsequence $b_n\to x$. Since $A$ and $B$ are closed sets, $x\in A\cap B$, a contradiction. Hence $A\cap B\ne \varnothing$.

The sum is nonempty because $A$ and $B$ are nonempty. It is a bounded set: if $|a|\le M_A$ and $|b|\le M_B$, then $|a+b|\le M_A+M_B$.

To prove that it is a closed set, let $x_j=a_j+b_j\in A+B$ be a convergent sequence with $x_j\to x$. By the Heine-Borel theorem, $A$ is a compact set, so $(a_j)$ has a convergent subsequence $a_{j_r}\to a\in A$. ThenSince $B$ is closed, $x-a\in B$, and hence $x=a+(x-a)\in A+B$. Therefore $A+B$ is nonempty, closed and bounded, so $A+B\in \mathcal{K}$.

Every $K_n$ is contained in the bounded set $K_1$, so the intersection $K={\bigcap }_{n\ge 1}{K}_n$ is bounded. It is closed because an arbitrary intersection of closed sets is closed.

It remains to prove that $K$ is nonempty. Choose $x_n\in K_n$. The set $K_1$ is a compact set, so some subsequence $x_{n_j}$ converges to a point $x\in K_1$. For each fixed $m$, all sufficiently late terms satisfy $n_j\ge m$ and hence $x_{n_j}\in K_{n_j}\subseteq K_m$. Since $K_m$ is closed, $x\in K_m$. This holds for every $m$, so $x\in K$. Thus $K\in \mathcal{K}$.

The condition $D{f}_p\ne 0$ says that $0$ is a regular value of $f$. The regular level set theorem therefore makes $f^{-1}(0)$ a smooth submanifold of codimension one in ${\mathbb{R}}^3$, hence a smooth surface.

Forthe gradient isOn $F^{-1}(0)$, $x^2+y^2=\cosh (z^2)\ge 1$, so $x$ and $y$ cannot both vanish. Thus $\nabla F\ne 0$ there, and the given set is a smooth surface.

Not every smooth surface in ${\mathbb{R}}^3$ is a global zero set. Every set $f^{-1}(0)$ is closed because $f$ is continuous, whereas the open unit diskis a smooth surface but is not closed in ${\mathbb{R}}^3$.

For $n\ge 0$, putEach $S_n$ is closed, the hypothesis gives $\mathbb{R}={\bigcup }_n{S}_n$, and $E_n\subset E_{n+1}$ because a function that vanishes on an open set has derivative zero there.

The open setis dense. Indeed, inside any nonempty open interval choose a nondegenerate closed subinterval $J$. The complete metric space $J$ is covered by the closed sets $J\cap S_n$, so the Baire's theorem makes one of them contain a relative open interval, which lies in some $E_n$.

On each connected component $I$ of $\Omega$, the function $F$ is one polynomial. To see this, any compact subinterval $K\subset I$ is covered by the increasing family $(E_n)$. A finite subcover therefore gives $K\subset E_N$ for some $N$, and $F^{(N)}=0$ on $K$. Hence $F$ is polynomial on $K$; overlapping compact intervals force these polynomials to agree throughout $I$.

Let $D=\mathbb{R}\setminus \Omega$. This closed set has no isolated points. Otherwise, on the two sides of an isolated point $x$, $F$ would equal polynomials. Smoothness makes all their one-sided derivatives agree at $x$, so the two polynomials are identical and extend across $x$. Some derivative would then vanish on a neighborhood of $x$, contrary to $x\in D$.

Suppose that $D$ is nonempty. It is complete andis a countable closed cover. A second application of Baire's theorem gives an index $k$ and an open interval $U_0$ such thatChoose $x_0\in D\cap U_0$ and a bounded open interval $U$ centred at $x_0$ whose closure lies in $U_0$.

Because $D$ has no isolated points, every $x\in D\cap U$ is approached by distinct points of $D\cap U$. Difference quotients first give $F^{(k+1)}(x)=0$, and induction gives

No component of $\Omega$ meeting $U$ can cross $x_0$, so it has an endpoint in $D\cap U$. If its polynomial had degree $d\ge k$, then its $d$th derivative would approach a nonzero constant at that endpoint, contradicting the preceding display. Its degree is therefore less than $k$, so $F^{(k)}$ also vanishes on the part of that component in $U$. We conclude that $F^{(k)}=0$ throughout $U$, which says $U\subset E_k\subset \Omega$ and contradicts $D\cap U\ne \varnothing$.

A sequentially closed set is one that contains every limit of its convergent sequences. This is weaker than being a closed set in a general topological space, but the two conditions agree in metric spaces.