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In this context a nowhere locally homogeneous metric, also called a bumpy metric, is a smooth Riemannian metric for which no two distinct nonempty open subsets are isometric with their induced metrics. Equivalently every local isometry between open subsets is the identity wherever defined: a nonidentity local isometry sends some point $x$ to a distinct point, and restriction to sufficiently small disjoint neighborhoods would violate the first formulation. This is the local-isometry meaning of the terminology here; degeneracy of periodic geodesics is a different use of the word bumpy.

Sunada's local isometry lemma states that on a compact smooth manifold without boundary of dimension $m\ge 2$ the nowhere locally homogeneous metrics contain a residual set in the space of smooth Riemannian metrics with its $C^{\infty }$ topology. In particular they are dense, by the Baire category theorem. A residual set is a countable intersection of open dense sets. The dimension assumption matters: every one-dimensional Riemannian metric is locally $d{s}^2$ in arclength coordinates and has local translations.

The jet bundle of maps $J^k(M,N)$ consists of equivalence classes $j_x^{k}f$ of smooth maps near $x$, where two maps agree to order $k$ at $x$ in coordinate charts. Its projection to $M\times N$ sends $j_x^{k}f$ to $(x,f(x))$. For a multi-index $\alpha$ in $m$ variables there are $\left(\genfrac{}{}{0ex}{}{m+r-1}{r}\right)$ derivatives of order $r$. Thus a coordinate chart consists of the source point, the target point, and $n$ coefficients for every derivative order $1$ through $k$. Summing the counts givesThis includes $J^0(M,N)=M\times N$. The fibre over $(x,y)$ is the space of truncated Taylor maps with fixed constant term $y$, locally modeled onFor $k=1$ its identification with $\mathrm{Hom}(T_{x}M,T_{y}N)$ is canonical. For higher $k$, changes of target coordinates mix derivatives of different orders, so this is a coordinate or connection-dependent description, not a canonical vector-bundle identification. The truncation $J^k\to J^{k-1}$ is an affine bundle modeled on the pullback of ${\mathrm{Sym}}^k(T^{\ast }M)\otimes TN$ over $M\times N$.

For the density assertion put $A={\overline{U}}_i$ and $B={\overline{U}}_j$. Their closures must be distinct. If $A=B$, the identity is an isometry for every Riemannian metric, and the requested complement is empty. Under the intended distinct-domain assumption, the smooth closed-ball hypothesis makes $A$ and $B$ regular closed domains, equal to the closures of their interiors. After interchanging them if necessary, there is a nonempty open set $W$ with compact closure in $\mathrm{int}(A)\setminus B$. Otherwise each interior would be contained in the other closure, forcing $A=B$.

Take any smooth Riemannian metric $g$. If ${\mathrm{vol}}_g(A)\ne {\mathrm{vol}}_g(B)$, it already lies outside ${\mathcal{S}}_{ij}$, because an isometry preserves the Riemannian volume form. If the volumes agree, choose a nonzero nonnegative smooth bump function $\eta$ supported in $W$ and setThese are positive definite Riemannian metrics, they agree with $g$ on $B$, and $g_t\to g$ in $C^{\infty }$ as $t\downarrow 0$, since every derivative of $g_t-g$ is $t$ times a fixed compactly supported smooth tensor. Their Riemannian volume forms satisfyHence ${\mathrm{vol}}_{g_t}(B)={\mathrm{vol}}_g(B)$ whereas ${\mathrm{vol}}_{g_t}(A)>{\mathrm{vol}}_g(A)$ for every $t>0$. In particularThe two domains cannot be isometric for $g_t$. Every $C^{\infty }$ neighborhood of $g$ therefore meets the complement of ${\mathcal{S}}_{ij}$, provingThis localized volume perturbation proves the requested density even when the two domains overlap, and works in every positive dimension. It does not by itself prove the stronger residual local-isometry statement: fixed isometric closures and arbitrary isometric open subsets are different conditions.

True. We prove weak-star topology on an entire infinite-dimensional Banach dual is not metrizable. Suppose instead that $E^{\prime }$ has a countable local base $(W_n)$ at zero. Choose a basic weak-star topology neighbourhood $U_n\subseteq W_n$, with its conditions involving a finite set $S_n\subseteq E$. The $U_n$ still form a local base.

For any $x\in E$, the set $\{\varphi :|\varphi (x)|<1\}$ is a neighbourhood of zero, so some $U_n$ is contained in it. Every functional annihilating $S_n$, and every scalar multiple of that functional, belongs to $U_n$. Consequently every such functional also annihilates $x$. This forcesIndeed a finite-dimensional span is norm closed, and the Hahn-Banach theorem supplies a bounded linear functional vanishing on it and nonzero at any point outside it.

It follows that $E={\bigcup }_n\mathrm{span}{S}_n$. Each span is a proper finite-dimensional closed vector subspace and has empty interior in the infinite-dimensional Banach space $E$. This contradicts the Baire category theorem. HenceThe uniform norm bound that made the metric work on $B^{\prime }$ is absent on the whole dual. The answers to (i), (ii), (iii) and (iv) are therefore all true.

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.

True. Apply the preceding Baire category theorem argument simultaneously to $\alpha =1/j$, $j\ge 1$. The union of all the corresponding $E_{m,j}$ is still a countable union of closed sets with empty interior, so its complement is nonempty. For any $\alpha >0$, choose $j$ with $1/j\le \alpha$. For $0<|h|\le 1$,The right-hand limsup is infinite at every $x$, proving the assertion for every positive exponent.

Work in the complete Banach space $C([0,1])$ with the uniform norm. Each $E_m$ is closed. Indeed if $f_j\to f$ uniformly and $x_j$ witnesses membership in $E_m$, pass to a subsequence with $x_j\to x$. For each $y\in [0,1]$, uniform convergence and continuity giveThus $x$ witnesses membership for $f$.

Each $E_m$ has empty interior. Approximate an arbitrary $f$ uniformly by a piecewise-linear function $q$ on a uniform mesh, and let $D$ bound the absolute slopes of $q$. On a sufficiently fine uniform refinement of mesh $\delta$, add a triangular oscillation of fixed small amplitude $b$, alternating between $b$ and $-b$ at successive vertices. The resulting function $g$ is still as close to $f$ as desired, and every segment has absolute slope at least $2b/\delta -D$. For any $x$, choose the farther endpoint of a segment containing $x$. Its distance $h$ is between $\delta /2$ and $\delta$, sowhen $\delta$ is small. Therefore $g\notin E_m$.

By the Baire category theorem, the complement of ${\bigcup }_m{E}_m$ is dense and nonempty. If the local limsup in the question were finite at some $x$, its nearby differences would satisfy a finite bound. The remaining differences, with $|h|\ge \delta >0$, are bounded by $2\Vert f{\Vert }_{\infty }{\delta }^{-\alpha }$. Combining the bounds would put $f$ in some $E_m$, a contradiction. Hence the required limsup is infinite at every point, including one-sided approaches at the endpoints.

The Baire category theorem in the needed version states that a nonempty complete metric space cannot be a countable union of closed sets with empty interior. To prove it, suppose $F_1,F_2,\dots$ are such sets and start with a nonempty open set $U$. Inductively choose closed balls ${\overline{B}}_j$ of positive radius at most $2^{-j}$, with ${\overline{B}}_1\subset U\setminus F_1$ and ${\overline{B}}_{j+1}\subset B_j\setminus F_{j+1}$. This is possible because each complement is open and dense. The centres form a Cauchy sequence; completeness supplies a limit in every closed ball. It lies in $U$ and outside every $F_j$, proving the theorem.

Here the sequence spaces are over the real scalars, as required by the Euclidean-plane assertion. Suppose an isometry $\phi :{\mathbb{R}}^2\to c_0$ existed. On the compact complete space $S^1\times S^1$, the setsare closed. They cover that space: for a nonzero sequence in $c_0$, the supremum norm is attained at a coordinate; for $x=y$ every coordinate works. By the Baire category theorem, one $F_n$ contains a product of open arcs. Shrink the arcs to be disjoint. The real coordinate difference is then continuous and nonzero, so its sign is constant on the product. Thus the chord distance has the separated formIn particular its alternating sum on every four-point rectangle is zero. Parametrize the arcs by angles $s,t$ with $0<t-s<2\pi$. Their chord distance is $D(s,t)=2\sin ((t-s)/2)$, whose mixed derivative isThe zero rectangle identity would force this mixed derivative to vanish, a contradiction. Therefore no such isometry exists.

For a real vector $x\in {\ell }_1^n$, list its signed sums over all $2^n$ choices $\epsilon \in \{-1,1{\}}^n$, followed by infinitely many zero coordinates:This is linear, has finite support and lies in $c_0$. Choosing the signs of the components gives ${\max }_{\epsilon }|\epsilon \cdot x|={\sum }_j|x_j|$. Thus $\boxed{\Vert Tx{\Vert }_{\infty }=\Vert x{\Vert }_1}$, the required linear isometry.

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.

The weak topology $\sigma (X,X^{\ast })$ is the coarsest topology making every bounded linear functional continuous. A neighbourhood basis at $x$ consists of sets $\{y:|f_j(y-x)|<ϵ,1\le j\le m\}$ for finite families in $X^{\ast }$.

Mazur theorem says that for any convex subset $C$ of a real or complex normed vector space,The norm closure is convex. If $x$ is outside it, the Hahn-Banach separation theorem gives a bounded linear functional $f$ and a real $a$ with $\mathrm{Re}f(x)>a\ge {\sup }_{y\in C}\mathrm{Re}f(y)$. In the real case omit the real part. Thus $x$ has a weak neighbourhood missing $C$, proving that the weak closure lies in the norm closure. The other inclusion follows because the weak topology is weaker than the norm topology. In the complex case real separation is converted to a complex functional by $f(z)=h(z)-ih(iz)$.

The sequential formulation, Mazur lemma, follows as well. If $x_n\rightharpoonup x$, then $x$ lies in the weak closure of each tail and hence in the norm closure of its convex hull. Select a finite convex combination of the $n$th tail at norm distance less than $1/n$ from $x$.

A weakly bounded set $D$ satisfies ${\sup }_{x\in D}|f(x)|<\infty$ for every $f\in X^{\ast }$. Regard $J(D)$ as a pointwise bounded family of functionals on $X^{\ast }$. The completeness of the dual space holds even if $X$ is incomplete: a norm-Cauchy sequence of functionals has a pointwise bounded linear limit and then converges uniformly on the unit ball. The Uniform boundedness principle therefore gives ${\sup }_{x\in D}\Vert Jx\Vert <\infty$, and $\Vert Jx\Vert =\Vert x\Vert$ proves norm boundedness.

One can see the precise Baire argument here. The closed sets $F_n=\{f\in X^{\ast }:{\sup }_{x\in D}|f(x)|\le n\}$ cover the Banach space $X^{\ast }$. The Baire category theorem makes some $F_n$ contain a ball $f_0+r{B}_{X^{\ast }}$, after shrinking the radius. For $\Vert g\Vert \le r$, both $f_0$ and $f_0+g$ lie in $F_n$, so ${\sup }_D|g(x)|\le 2n$. Scaling and the dual norm formula yield ${\sup }_D\Vert x\Vert \le 2n/r$. A weakly compact set is weakly bounded because each functional has compact, hence bounded, image. Consequently it is norm bounded.

For a Banach space, let $J:X\to X^{\ast \ast }$ be its canonical isometry. We use two explicitly stated weak-star facts: Banach-Alaoglu theorem makes a dual unit ball weak-star compact, and Goldstine theorem makes $J(B_X)$ weak-star dense in $B_{X^{\ast \ast }}$. The weak topology on $X$ is carried by $J$ to $\sigma (X^{\ast \ast },X^{\ast })$ on its image, since their coordinates are the same evaluations $f(x)$.

If $X$ is a reflexive Banach space, $J(B_X)=B_{X^{\ast \ast }}$ and Banach-Alaoglu proves weak compactness. Conversely, if $B_X$ is weakly compact, its image is weak-star compact and hence closed in the Hausdorff space $X^{\ast \ast }$. Goldstine density then forces $J(B_X)=B_{X^{\ast \ast }}$, which implies $JX=X^{\ast \ast }$. This proves the weak compactness characterization of reflexivity:Finally let $K\subset X$ be weakly compact and let $(f_n)\subset B_{X^{\ast }}$ separate points. If $K$ is empty there is nothing to prove; otherwise defineThe summands are bounded by $2^{-n}$ and separation makes $d(x,y)=0$ only for $x=y$. The coordinate maps are weakly continuous, so the series, being uniformly convergent, makes the identity from weak $K$ to metric $K$ continuous. A continuous bijection from a compact space to a Hausdorff space is a homeomorphism. Thus $d$ induces precisely the weak topology on $K$, the countable separating family metrizes a weakly compact set result. No norm-density of the separating family in $X^{\ast }$ is claimed or required.

The weak-star topology $\sigma (X^{\ast },X)$ is the coarsest topology making $f\mapsto f(x)$ continuous for every $x\in X$. Its basic neighbourhoods of zero impose finitely many inequalities $|f(x_j)|<ϵ$.

Suppose first that zero had a countable neighbourhood base $(U_n)$ in the entire dual. Choose a basic neighbourhood $V_n\subset U_n$ controlled by a finite set $F_n\subset X$, and let $E_n=\mathrm{span}(F_1\cup \cdots \cup F_n)$. Given $x\in X$, the neighbourhood $W_x=\{f:|f(x)|<1\}$ contains some $U_n$, hence $V_n$. If $x\notin \mathrm{span}{F}_n$, the Hahn-Banach theorem supplies a bounded linear functional $h$ which vanishes on $F_n$ but has $h(x)\ne 0$. Indeed, the finite-dimensional span is closed, and the functional taking $e+tx$ to $t$ on its sum with $\mathbb{F}x$ is bounded because $\mathrm{dist}(x,\mathrm{span}{F}_n)>0$. Every scalar multiple of $h$ lies in $V_n$, contradicting $V_n\subset W_x$. Thus $x\in E_n$, and $X={\bigcup }_n{E}_n$ has countable Hamel dimension.

If $X$ is infinite-dimensional, every finite-dimensional $E_n$ is closed and has empty interior. Completeness rules out their union. For clarity, the Baire category theorem needed here has a direct nested-ball proof: inside a starting open ball choose a closed ball missing $E_1$, then inside its interior choose a closed ball missing $E_2$, and continue with positive radii tending to zero. The centres are Cauchy; their limit belongs to every ball and to none of the $E_n$, a contradiction. A metric topology would have a countable ball base at zero. HenceThis is the weak-star topology on an entire infinite-dimensional Banach dual is not metrizable result. It does not contradict weak-star metrizability of the dual ball for a separable predual: that assertion concerns a bounded subset.

The Banach-Alaoglu theorem states that $B_{X^{\ast }}$ is compact for $\sigma (X^{\ast },X)$ for every normed vector space $X$, whether complete or not. Embed it in the product ${\prod }_{x\in X}\{a\in \mathbb{F}:|a|\le \Vert x\Vert \}$ by its evaluations. Each factor is compact, so Tychonoff theorem makes the product compact Hausdorff. The equations expressing additivity and scalar homogeneity define a closed subset of the product. Its elements are exactly the bounded linear functionals of norm at most one, since the coordinate bounds give $|f(x)|\le \Vert x\Vert$. The induced product topology is precisely weak-star, proving the theorem.

The theorem printed as “Goldstein” is the standard Goldstine theorem: $J(B_X)$ is weak-star dense in $B_{X^{\ast \ast }}$ for the canonical embedding into the bidual. To prove it, fix $\Phi \in B_{X^{\ast \ast }}$ and $f_1,\dots ,f_m\in X^{\ast }$. The set $S=\{(f_1(x),\dots ,f_m(x)):x\in B_X\}$ is convex. If $b=(\Phi (f_1),\dots ,\Phi (f_m))$ were outside its closure, finite-dimensional Hahn-Banach separation theorem would supply coefficients $a_j$ such that, for $f={\sum }_j{a}_j{f}_j$,The real case omits real parts; in the complex case every real-linear separating functional on ${\mathbb{C}}^m$ has the displayed form. The supremum equals the dual norm by multiplying vectors by signs or unimodular scalars. The strict inequality contradicts $\Vert \Phi \Vert \le 1$. Thus $b\in \overline{S}$, which is exactly approximation on every prescribed finite family of weak-star coordinates.

Now take $K=B_{X^{\ast }}$ with its compact Hausdorff weak-star topology. For each $x\in X$, the evaluation $\hat{x}:f\mapsto f(x)$ is continuous on $K$, and Hahn-Banach theorem gives $\Vert \hat{x}{\Vert }_{\infty }=\Vert x\Vert$. Thereforeis the isometric evaluation embedding into continuous functions on a compact dual ball, over the same scalar field as $X$.

For the final sequential assertions, choose a norm-dense sequence $(f_j)$ in the separable dual $X^{\ast }$ and suppose $\Vert x_n\Vert \le M$. Each bounded scalar sequence $f_j(x_n)$ has a convergent subsequence: divide a bounding interval or square into finitely many smaller closed pieces, repeatedly retain one containing infinitely many terms, and let the diameters tend to zero. Successive extraction for $j=1,2,\dots$, followed by a diagonal subsequence $(y_n)$, makes every $f_j(y_n)$ converge.

If $M=0$ the conclusion is immediate. Otherwise, for any $f\in X^{\ast }$ and $ϵ>0$, choose $j$ with $\Vert f-f_j\Vert <ϵ/(4M)$. For sufficiently large $n,m$,Thus $\phi (f)={\lim }_{n}f(y_n)$ exists for every $f$. Taking limits proves linearity, and $|\phi (f)|\le M\Vert f\Vert$ provesThis diagonal argument includes the continuity estimate rather than appealing to a sequential compactness theorem.

If $X$ is not reflexive, choose $\Phi \in B_{X^{\ast \ast }}\setminus JX$, scaling a bidual element outside $JX$ if necessary. The Goldstine theorem proved above supplies $x_n\in B_X$ with $|f_j(x_n)-\Phi (f_j)|<1/n$ for $j\le n$. Norm density and the bound $\Vert J{x}_n-\Phi \Vert \le 2$ extend this convergence to every $f\in X^{\ast }$. Hence $J{x}_n\to \Phi$ weak-star. Every subsequence has the same coordinate limits, so a weakly convergent subsequence with limit $x\in X$ would give $Jx=\Phi$, a contradiction. This sequential Goldstine approximation for a separable dual provesThe construction uses the Goldstine proof already given, not an unproved weak sequential compactness result.

The Baire category theorem says that in a nonempty complete metric space, every countable intersection of open dense sets is dense. To prove it, start in an arbitrary nonempty open set $O$ and let $G_n$ be open dense. Inductively choose closed balls $\overline{B}(x_n,r_n)$ with $0<r_n\le 2^{-n}$, contained in $G_n$ and in the interior of the previous ball, with the first also contained in $O$. Density provides a point at each stage, and openness permits the smaller closed ball. The centers are a Cauchy sequence; completeness supplies a limit lying in every closed ball, hence in $O\cap {\bigcap }_n{G}_n$. The proof also works at isolated points, where sufficiently small balls are singletons.

The incomplete metric space $\mathbb{Q}$ is a counterexample: it is the countable union of its closed nowhere dense singletons, so the complements are open dense with empty intersection.

For the sequence in the problem, putEach $E_N$ is closed by continuity. Pointwise convergence makes each point's values a Cauchy sequence, so $[0,1]={\bigcup }_N{E}_N$. By the Baire category theorem, one $E_N$ has nonempty relative interior, which contains a nonempty open interval $J$ in $(0,1)$. This proves the required common tail bound.

If continuous $g_n$ converged pointwise to the rationality indicator, take $ϵ=1/4$ and let $m\to \infty$ in this bound. On $J$, $|g_n-g|\le 1/4$ for $n\ge N$. Hence a fixed continuous $g_N$ is at least $3/4$ at every rational point and at most $1/4$ at every irrational point of $J$. The density of both sets contradicts continuity. Thus the rationality indicator is not a pointwise limit of continuous functions.

Nevertheless $h_n(x)=x^n$ are continuous and converge pointwise to $h(x)=0$ for $x<1$, $h(1)=1$. This limit is discontinuous at $1$: a pointwise limit of continuous functions can be discontinuous.

One form of the Baire category theorem is: a nonempty complete metric space cannot be a countable union of closed nowhere dense subsets. Equivalently, a countable intersection of open dense subsets is dense. A Banach space is a complete metric space for its norm metric, so this form applies directly to the next part.

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.

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.

For each $n\ge 0$, defineEvery $f^{(n)}$ is an entire function, hence continuous, so $E_n$ is closed. The hypothesis saysThe complex plane is a complete metric space. The Baire category theorem therefore implies that some $E_N$ has nonempty interior. Thus $f^{(N)}$ vanishes on an open set, and the identity theorem giveson all of $\mathbb{C}$. If $N=0$, then $f=0$; otherwise the power series of $f$ has no term of degree $N$ or higher. Hence $f$ is a polynomial of degree at most $N-1$. This is the pointwise vanishing derivative criterion for a polynomial.

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 $q<p$.