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A set $N$ is completely Ramsey-null when every Ellentuck topology neighborhood $[s,A]$ has a refinement $[s,B]$ disjoint from $N$, with the same finite stem $s$. Assuming open Ellentuck sets are completely Ramsey, every star-nowhere dense set is completely Ramsey-null: apply the open-set property to the dense complement of its closure. The Ellentuck meagre-set fusion lemma shows that countable unions retain this avoidance property.

A meagre, or first-category, set is a countable union of nowhere dense sets.

If an entire function $f$ is not a polynomial, then there is a point $z_0\in \mathbb{C}$ for which $f^{(n)}(z_0)\ne 0$ for every nonnegative integer $n$. Consequently every coefficient in the Taylor series of $f$ about $z_0$ is nonzero.

For each $n$, the function $f^{(n)}$ is not identically zero, since otherwise $f$ would be a polynomial. Its zero set is closed and has empty interior by the identity theorem, hence is a nowhere dense set. The Baire category theorem says that the complete metric space $\mathbb{C}$ cannot be the union of these countably many zero sets.

Set $S_n=\{x:f^{(n)}(x)=0\}$ and $E_n=\mathrm{int}{S}_n$. The sets $S_n$ are closed, $E_n\subset E_{n+1}$, and Baire's theorem on each interval shows that $\Omega ={\bigcup }_n{E}_n$ is dense. On every connected component of $\Omega$, compactness and the increasing cover by the $E_n$ show that one derivative vanishes locally with a uniform finite order, so $f$ agrees there with one polynomial.

If $F=\mathbb{R}\setminus \Omega$ were nonempty, it would be closed and have no isolated points: polynomial pieces on both sides of an isolated point would have matching derivatives of every order and hence join into one polynomial piece. Applying Baire's theorem to the cover $F={\bigcup }_n(F\cap S_n)$ gives an interval $U$ and an index $k$ for which $F\cap U$ is nonempty and contained in $S_k$. Because $F$ has no isolated points, difference quotients show that every derivative of order at least $k$ vanishes on $F\cap U$. Each polynomial component of $\Omega$ meeting $U$ has an endpoint in $F\cap U$, so its degree is less than $k$. Thus $f^{(k)}=0$ throughout $U$, contradicting $F\cap U\ne \varnothing$. Hence $\Omega =\mathbb{R}$, and its sole connected component carries one polynomial.

Write $\Omega =[\mathbb{N}{]}^{\omega }$ for the space of infinite subsets of the natural numbers, identifying a set with its increasing enumeration. If $s$ is finite and $A$ is infinite with every element of $A$ above $\max s$, setFor the empty stem there is no lower-bound restriction. These sets form the basis of the Ellentuck topology, also called the star topology. The stem $s$ is fixed, while the infinite tail may be thinned.

A completely Ramsey set $E\subseteq \Omega$ is one for which every $[s,A]$ admits an infinite $B\subseteq A$ with $[s,B]\subseteq E$ or $[s,B]\cap E=\varnothing$. This keeps the same stem. A Ramsey set of infinite subsets requires that every infinite $A$ admit an infinite $B\subseteq A$ with a homogeneous empty-stem cone $[B{]}^{\omega }$. It does not require preserving a nonempty finite stem.

A set is a star-Baire set if it differs from a star-open set by a star-meagre set, equivalently $E△U$ is star-meagre for some star-open $U$. A nowhere dense set has closure with empty interior, and a meagre set is a countable union of such sets, with both notions interpreted in the specified topology.

For a non-Ramsey example, put $M=\{2,3,\dots \}$ and well-order all infinite subsets of $M$ as $(A_{\alpha }{)}_{\alpha <\mathfrak{c}}$, where $\mathfrak{c}=2^{{\aleph }_0}$. Recursively choose two fresh points $X_{\alpha },Y_{\alpha }\in [A_{\alpha }{]}^{\omega }$, not used at earlier stages. This is possible because each cone has cardinality $\mathfrak{c}$ and fewer than $\mathfrak{c}$ points have been used before any stage. Letkeeping all the $Y_{\alpha }$ outside $C$. Every infinite-subset cone on $M$ meets both $C$ and its complement. Thus $C$ is not Ramsey, even when regarded as a subset of $\Omega$: any infinite set can first be thinned to avoid $1$. This uses the axiom of choice, rather than claiming a Borel counterexample.

Now setEvery infinite $A$ has an infinite subset $B$ avoiding $1$, and then $[B{]}^{\omega }\cap E=\varnothing$. Hence $E$ is Ramsey. But no stem-preserving thinning of $[\{1\},M]$ is homogeneous for $E$, because $C$ splits every tail cone. Hence $E$ is not completely Ramsey.

To prove the topological equivalence, we first establish the Ellentuck meagre-set fusion lemma: every star-meagre set can be avoided in a refinement $[s,B]$ with the original stem. Call a set completely Ramsey-null if this avoidance holds in every $[s,A]$.

If $N$ is star-nowhere dense, then $U=\Omega \setminus \overline{N}$ is star-open and dense. By the granted complete-Ramsey property of star-open sets, every $[s,A]$ has a refinement $[s,B]$ either contained in $U$ or disjoint from $U$. The latter alternative would put a nonempty open set inside $\overline{N}$, contradicting density of $U$. Thus $N$ is completely Ramsey-null.

For a countable union ${\bigcup }_j{N}_j$ of completely Ramsey-null sets, use fusion, taking care of every possible finite stem. Starting with $A_0=A$, at stage $j$ choose $b_j\in A_{j-1}$ above all previous choices. From the part of $A_{j-1}$ above $b_j$, successively thin an infinite tail for each of the finitely many subsets $t\subseteq \{b_1,\dots ,b_j\}$ so thatEach thinning keeps the stem $s\cup t$ fixed; subsequent thinnings preserve earlier avoidance. Let $B=\{b_1,b_2,\dots \}$. For any $X\in [s,B]$ and fixed $j$, put $t=(X\setminus s)\cap \{b_1,\dots ,b_j\}$. The rest of $X$ lies in $A_j$, because all later selected points do. Therefore $X\in [s\cup t,A_j]$ and $X\notin N_j$. Since $j$ was arbitrary,This proves the fusion lemma. Considering all subsets $t$, not merely the single selected prefix, is essential: elements of $[s,B]$ need not use every $b_i$.

Suppose first that $E$ is star-Baire, with $E△U$ star-meagre and $U$ star-open. Thin $[s,A]$ to $[s,B]$ avoiding the meagre difference by the lemma. Then apply the granted complete-Ramsey property of $U$ to find $[s,D]$ homogeneous for $U$, with $D\subseteq B$. On this refinement $E$ and $U$ agree, so it is homogeneous for $E$. Thus $E$ is completely Ramsey.

Conversely, let $E$ be completely Ramsey and let $U$ be the union of all basic neighborhoods wholly contained in $E$, namely its star-interior. Every $[s,A]$ has a homogeneous refinement. If that refinement lies in $E$, it lies in $U$; if it avoids $E$, it also avoids $E\setminus U$. Consequently every basic neighborhood contains a nonempty open refinement disjoint from $E\setminus U$. Since such a refinement is also disjoint from the closure of $E\setminus U$, this difference is star-nowhere dense. Thus $E△U=E\setminus U$ is star-meagre, and $E$ is star-Baire. We have provedFinally, $\Omega$ is not star-meagre: otherwise apply the fusion lemma to its purported meagre cover inside any nonempty basic neighborhood, obtaining a nonempty $[s,B]$ disjoint from $\Omega$, a contradiction.

The printed paper does not define $\tau$. In the coarser Ramsey cone topology, whose basic open sets are $[A{]}^{\omega }$ without finite stems, $\Omega$ is $\tau$-meagre. To prove this, let $D_j=\{X:\min X=j\}$. Its cone-topology closure is $H_j=\{X:j\in X\}$: every cone neighborhood of a point containing $j$ has a subset with least element $j$, whereas if $j\notin X$ the neighborhood $[X{]}^{\omega }$ avoids $D_j$. No nonempty cone lies in $H_j$, since its infinite ground set can be thinned to remove $j$. Therefore each $D_j$ is nowhere dense, but $\Omega ={\bigcup }_{j\ge 1}{D}_j$. This proves meagreness of the Ramsey cone topology and exhibits the contrast with the star topology.

If $\tau$ instead denotes the ordinary topology on infinite subsets, with basic cylinders $[s]=\{X:s\text{ is an initial segment of }X\}$, the answer is not $\tau$-meagre. Indeed, given countably many nowhere dense sets, extend a finite increasing prefix successively so that its cylinder avoids the closure of the next set. Make the prefix longer at each step. Their union is an infinite increasing enumeration belonging to every chosen cylinder and avoiding the whole proposed cover. The same construction starts inside any nonempty cylinder, so this alternative product topology is also a Baire space. The two conventions have different answers, so the distinction must be made explicitly rather than inferred from the symbol alone.

The construction gives a Lusin set $\boxed{A=\{x_{\alpha }:\alpha <{\omega }_1\}}$ with all $x_{\alpha }$ distinct.

Every meagre set is contained in a meagre $F_{\sigma }$ set: replace each nowhere dense set in a countable covering by its closure. There are at most $2^{{\aleph }_0}$ such covers, since each closed subset of $\mathbb{R}$ is determined by a countable rational basis for its open complement, and a countable sequence of such codes is again coded by a real. Under the Continuum hypothesis, enumerate all meagre $F_{\sigma }$ sets as $(M_{\alpha }{)}_{\alpha <{\omega }_1}$, allowing repetitions.

By transfinite recursion, chooseFor each $\alpha <{\omega }_1$, the excluded union is a meagre set: it is a countable union of meagre sets and singletons. The Baire category theorem ensures that it does not cover $\mathbb{R}$, so the choice is possible. The resulting $A$ is uncountable. If $M$ is meagre, choose $\gamma$ with $M\subseteq M_{\gamma }$. Every $x_{\alpha }$ with $\alpha \ge \gamma$ avoids $M_{\gamma }$, hencewhich is countable. Thus $A$ has precisely the Lusin set property. Enumerating covers rather than all meagre sets avoids an unjustified claim that there are only continuum many arbitrary meagre subsets.

Because $Y$ is closed, it is a nowhere dense set exactly when its interior is empty. If a ball about $y\in Y$ lay inside $Y$, translating it by $-y$ would put a ball about $0$ inside $Y$. Every vector of $X$ has a small scalar multiple in that ball, and closure under scalar multiplication would force $Y=X$. This contradiction proves a proper closed linear subspace is nowhere dense.

Suppose $(x_k{)}_{k\ge 1}$ were a countable Hamel basis. Put $E_n=\mathrm{span}(x_1,\dots ,x_n)$. Each $E_n$ is a finite-dimensional, hence closed, linear subspace of $X$. It is proper because $X$ is infinite dimensional, and therefore a nowhere dense set. A Hamel basis means that every vector is a finite linear combination, so $X={\bigcup }_n{E}_n$. This contradicts the Baire category theorem for the Banach space $X$. Hence an infinite-dimensional Banach space has no countable Hamel basis. The same proof excludes any countable algebraic spanning set.

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

One form of the Baire category theorem says that a nonempty complete metric space cannot be a countable union of closed nowhere dense sets.

For each nonnegative integer $n$, putBecause $f$ is an entire function, every derivative $f^{(n)}$ is continuous and $Z_n$ is closed. No $f^{(n)}$ is identically zero: if $f^{(n)}\equiv 0$ for some $n$, repeated integration shows that $f$ is a polynomial. By the identity theorem, the zero set of the nonzero holomorphic function $f^{(n)}$ has empty interior. Thus each $Z_n$ is nowhere dense.

The complex plane is complete, so the Baire category theorem givesChoose $z_0$ outside this union. Then $f^{(n)}(z_0)\ne 0$ for every $n$, and hence every coefficient $f^{(n)}(z_0)/n!$ in the Taylor series about $z_0$ is nonzero. Such a point $z_0$ therefore exists.

Suppose for contradiction that the vector space of univariate polynomials $P$ were complete in the given norm. For $n\ge 0$, letEach $P_n$ is finite-dimensional and therefore closed by the fact that a finite-dimensional subspace is closed. It is a proper linear subspace of $P$, so it has empty interior: if a linear subspace contained any open ball, translation and scalar multiplication would make it contain every vector. Hence every $P_n$ is a nowhere dense set.

But every polynomial has finite degree, soThis writes the supposed complete metric space $P$ as a countable union of nowhere-dense closed subsets, contradicting the Baire category theorem. Thus no norm can make $P$ complete. Equivalently, this is the countable-basis obstruction that a Banach space has uncountable Hamel dimension.

The Baire category theorem says that a complete metric space cannot be a countable union of nowhere dense sets, equivalently that every countable intersection of open dense sets is dense. To prove it, let $G_1,G_2,\dots$ be open dense subsets and let $V$ be nonempty and open. Choose nested closed ballsstarting inside $V\cap G_1$. The centres are a Cauchy sequence; completeness gives a limit lying in every ball, hence in $V\cap {\bigcap }_n{G}_n$. Thus the intersection is dense.

Choose a countable sequence $q_j\uparrow p$ with $1\le q_j<p$. Since ${\ell }^q\subset {\ell }^r$ for $q<r$, every ${\ell }^q$ with $q<p$ lies in some ${\ell }^{q_j}$. For integers $m\ge 1$, setPointwise convergence and Fatou lemma for series show that $E_{j,m}$ is closed in ${\ell }^p$. It has empty interior: inside any ${\ell }^p$ ball, add a sufficiently long finite block whose ${\ell }^p$ norm is tiny but whose ${\ell }^{q_j}$ norm is larger than $m$. Therefore every $E_{j,m}$ is nowhere dense. Since ${\ell }^p$ is a Banach space, Baire gives

An explicit element of the difference isIts $p$th powers form a convergent logarithmic series, whereas for every $q<p$ the powers contain $n^{-q/p}$ with exponent below one and their series diverges. Hence $x\in {\ell }^p$ but $x\notin {\ell }^q$ for every $q<p$.