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A compact metric space is second countable. To see this concretely, for each positive integer $r$ choose a finite cover by balls of radius $1/r$. Their centres, over all $r$, form a countable dense subset; balls centred there with positive rational radii form a countable basis of a topology. Enumerate its nonempty members as $(V_j{)}_{j\ge 1}$, repeating members if the basis is finite.

For each $j$, putEach $G_j$ is an open set, because all iterations of a map $f^n$ are continuous. It is dense: if $U$ is any nonempty open set, the hypothesis supplies $n\ge 1$ and $u\in U$ with $f^n(u)\in V_j$, so $u\in U\cap G_j$.

A compact metric space is a complete metric space, hence a Baire space. The Baire category theorem implies that $G={\bigcap }_{j\ge 1}{G}_j$ is dense, and in particular nonempty. For any $x\in G$, the forward orbit of $x$ meets every $V_j$, and therefore every nonempty open set. Its orbit closure is $X$. Thus the open-set intersection hypothesis implies point transitivity, and in fact the set of points with a dense forward orbit contains the dense intersection $G$.

The Baire category theorem states that a complete metric space is a Baire space: the intersection of countably many open dense sets is dense. Equivalently, a nonempty open subset cannot be a countable union of nowhere dense sets.

Let $U_n$ be open dense and $O$ any nonempty open set. Choose a closed ball $B_1$ of positive radius at most $1/2$ inside $O\cap U_1$. Inductively choose $B_{n+1}$ inside the interior of $B_n$ intersected with $U_{n+1}$, with positive radius at most $2^{-(n+1)}$. Density makes this intersection nonempty, and openness supplies a sufficiently small closed ball. The centres form a Cauchy sequence because all later centres lie in $B_n$, whose diameter tends to zero. Completeness gives a limit $x$. Since each ball is closed and contains all later centres, $x\in B_n$ for every $n$. Thus $x\in O\cap {\bigcap }_n{U}_n$, proving density. Applying this to complements of the closures of nowhere dense sets gives the equivalent formulation.

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.

A subspace is topologically complete if its subspace topology is induced by some complete metric. The compatible complete metric need not be the restriction of the displayed ambient metric. For example, an open interval is topologically complete although its usual metric is incomplete.

Suppose $\rho$ is a compatible complete metric on $Y$. For each $n\ge 1$ and $y\in Y$, choose an ambient open set $V_{n,y}$ containing $y$, contained in $B_d(y,1/n)$, and with $\rho$-diameter of $V_{n,y}\cap Y$ at most $1/n$. Such sets exist because the two topologies on $Y$ agree. Put $U_n={\bigcup }_{y\in Y}{V}_{n,y}$. Clearly $Y\subseteq {\bigcap }_n{U}_n$.

If $x\in {\bigcap }_n{U}_n$, choose $y_n\in Y$ with $x\in V_{n,y_n}$. Then $d(x,y_n)<1/n$, so $y_n\to x$ in the ambient metric. For fixed $m$, the point $x$ belongs to some open $V_{m,z}$, and eventually all $y_n$ belong to this set. Their pairwise $\rho$-distances are then at most $1/m$. Hence $(y_n)$ is $\rho$-Cauchy, and completeness gives a limit $y\in Y$. Compatibility gives $d(y_n,y)\to 0$, forcing $x=y$. ThusThis proves the G-delta criterion for topological completeness in the required direction.

The converse is true in a complete metric ambient space. If $Y={\bigcap }_n{U}_n$ with each $U_n$ open in complete $X$, define $a_n(y)=1/d(y,X\setminus U_n)$, omitting any term whose complement is empty. A compatible complete metric on $Y$ isContinuity of each $a_n$ and the uniformly small series tail show that $D$ has the original topology. A $D$-Cauchy sequence is $d$-Cauchy, hence converges to some $x\in X$. For each $n$, its real coordinates $a_n$ form a Cauchy sequence and remain bounded. Since distance to a closed set is continuous, $d(x,X\setminus U_n)>0$, so $x\in U_n$ for every $n$. The same coordinate convergence and series-tail argument give convergence in $D$. Thus $D$ is complete.

For the normed-space claim, embed $E$ densely in its norm completion $\hat{E}$. Topological completeness and the preceding criterion make $E$ a dense G-delta set in $\hat{E}$, hence comeagre. Every translate $x+E$ is also comeagre. The Baire category theorem in the complete space $\hat{E}$ implies that $E\cap (x+E)$ is nonempty. If $z=x+e$ lies in that intersection, then $x=z-e\in E$. This holds for every $x\in \hat{E}$, soThis is the comeagre subgroup completeness argument.

Finally let $B=\{\ell \in E^{\ast }:\Vert \ell \Vert <1\}$ with its weak-star topology. Put $r_n=1-1/(n+1)$ and $F_n=\{\ell \in B:\Vert \ell \Vert \le r_n\}$. The dual norm is the supremum of the continuous evaluation moduli on the unit ball of $E$, so every $F_n$ is relatively closed. Also $B={\bigcup }_n{F}_n$.

Each $F_n$ has empty relative interior. Given $\ell \in F_n$ and a basic weak-star neighborhood restricting finitely many evaluations on $x_1,\dots ,x_k$, the Hahn-Banach theorem gives a nonzero functional $h$ annihilating their finite-dimensional span, because $E$ is infinite-dimensional. All $\ell +th$ retain those evaluations. Their norm varies continuously with $t$ and becomes unbounded; choose $t$ so that $r_n<\Vert \ell +th\Vert <1$. This point lies in the neighborhood inside $B$ but outside $F_n$. Thus $B$ is a nonempty countable union of relatively closed nowhere dense sets and is not a Baire space. ConsequentlyThis weak-star open dual ball category obstruction works without assuming that $E$ is separable or that the weak-star ball is metrizable.

Part i makes $K_{\epsilon }^{\prime }$ a closed subset of the compact metrizable space $K$, so it is compact metrizable and therefore separable. Choose the stated dense sequence $(f_m)$ and, using part iii, the sequences $(g_{m,n})$ and $(y_{m,n})$. In constructing the universal weakly null sequence $(z_n)$ from part b, retain a tail of each $y_m$ and write $z_{j(m,n)}=y_{m,n}$ along the retained subsequence, where $j(m,n)\to \infty$.

Eachis weak-star closed. Since $z_n\rightharpoonup 0$, every fixed $f\in K$ belongs to some $A_N$, so $K={\bigcup }_N{A}_N$. The compact Hausdorff space $K$ is a Baire space, and the Baire category theorem implies that some $A_N$ has nonempty relative weak-star interior.

Suppose $K_{\epsilon }^{\prime }=K$. Choose a nonempty relatively open $O\subseteq A_N$ and then $f_m\in O$ by density. Since $g_{m,n}\stackrel{w^{\ast }}{\to }{f}_m$, eventually $g_{m,n}\in O$. For a sufficiently late retained index, also $j(m,n)\ge N$, and hencecontradicting $|g_{m,n}(y_{m,n})|>\epsilon /16$. Therefore the One-step Szlenk derivation for a separable dual gives $K_{\epsilon }^{\prime }⊊K$ whenever $K$ is nonempty.

A set containing a countable intersection of open dense subsets. The intersection itself is also called a residual set. In a Baire space it is dense. This topological notion of genericity does not by itself assert full measure.

For an infinite-dimensional normed space, the weak-star open unit ball of its dual is not a Baire space. The smaller closed norm balls are weak-star closed and have empty relative interior: a nonzero functional annihilating any finitely many tested vectors can move the norm beyond the smaller radius while remaining below one. Their countable union is the open norm ball. Thus this topology is not topologically complete, regardless of separability.