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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.

The weak topology of probability measures is the smallest topology making every map $\mu \mapsto \int hd\mu$ continuous for bounded continuous real $h$ on $X$. Define the bounded-Lipschitz metric byUsing the maximum of the two bounds instead of their sum changes the metric by at most a factor of two and gives the same topology. Symmetry and the triangle inequality follow from the supremum formula. Bounded Lipschitz functions distinguish probability measures by the open-set approximations below, so $\beta$ is a metric. On a separable metric space, $\beta$ induces the weak topology.

Here are the essential details. If $\beta ({\mu }_n,\mu )\to 0$, integrals of every bounded Lipschitz function converge after rescaling its Lipschitz bound. For an open set $A\ne X$, the functionsincrease to $1_A$; for $A=X$ take $h_m=1$. Thus $\mu (A)={\sup }_m\int h_{m}d\mu$. It follows both that $\mu \mapsto \mu (A)$ is lower semicontinuous for the weak topology, as a supremum of continuous maps, and that beta convergence impliesThis is the open-set Portmanteau criterion. Its closed-set equivalent, by complements, isEither family of inequalities is necessary and sufficient for weak convergence of probability measures. For sufficiency of the open inequalities, apply the layer-cake formula and Fatou's lemma to the open superlevel sets of a continuous function $0\le h\le 1$, obtaining $\mathrm{liminf}\int hd{\mu }_n\ge \int hd\mu$. Applying the same argument to $1-h$ gives the reverse bound. Scaling handles every bounded continuous function.

Conversely, weak convergence implies beta convergence. Choose a compact set $K$ of arbitrarily large $\mu$-mass, using tightness of a probability measure on a Polish space. The unit bounded-Lipschitz class is uniformly bounded and equicontinuous on $K$, so finitely many of its members approximate all others uniformly there. On the open $\delta$-neighborhood $K^{\delta }$, approximation errors increase by at most $2\delta$. The open-set inequality makes ${\mu }_n(K^{\delta })$ large for all sufficiently large $n$. Weak convergence for the finitely many selected functions, combined with these approximation errors and the small mass outside $K^{\delta }$, bounds the supremum defining $\beta$ by an arbitrarily small number. The same finite-test and open-mass bounds define weak neighborhoods, so the argument gives equality of the topologies, not only their convergent sequences. It does not first assume uniform tightness of the entire sequence.

The map $j=i_{\ast }$ is injective and continuous: for $h\in C(\tilde{X})$, $h\circ i$ is bounded continuous on $X$, and $\int hdj(\mu )=\int h\circ id\mu$. To prove inverse continuity on its image, suppose $j({\mu }_n)\to j(\mu )$. For every open $A\subset X$, there is an open $U\subset \tilde{X}$ with $i(A)=U\cap i(X)$. Hence ${\mu }_n(A)=j({\mu }_n)(U)$ and $\mu (A)=j(\mu )(U)$. The open-set criterion on $\tilde{X}$ now gives the criterion on $X$, so ${\mu }_n\to \mu$. Both spaces are metrizable, and this sequential argument provesThis is the probability pushforward embedding theorem. The printed claim needs the words “onto its image”: it is generally not onto all $P(\tilde{X})$, since a Dirac mass at a point outside $i(X)$ is not in its image. By Question 1, the completely metrizable subspace $i(X)$ is a Borel G-delta set, and

Now let $S$ be weakly closed and uniformly tight. Take any sequence $({\mu }_n)$ in $S$. The space $P(\tilde{X})$ is compact for the weak topology. For completeness, this compactness follows by choosing a countable uniformly dense family in $C(\tilde{X})$, extracting a diagonal subsequence of its bounded integrals, and extending the limits to a positive normalized functional on $C(\tilde{X})$; the Riesz-Markov-Kakutani representation theorem gives the limiting probability measure. Thus, along a subsequence, $j({\mu }_n)\to \lambda$.

For every $m$, choose compact $K_m\subset X$ with $\mu (K_m)\ge 1-2^{-m}$ for every $\mu \in S$. The set $i(K_m)$ is compact and closed in $\tilde{X}$, so the closed-set criterion givesTherefore $\lambda (i(X))=1$, and $\lambda =j(\mu )$ for some $\mu \in P(X)$. Inverse continuity gives ${\mu }_n\to \mu$, and closedness puts $\mu \in S$. Hence $S$ is sequentially compact, and metrizability makes it compact. This proves the required closed uniformly tight compactness criterion, rather than assuming it from Prokhorov's theorem.

Such a compact metrizable extension is always available: for a countable dense set $(x_k)$, the coordinates $d(x,x_k)/(1+d(x,x_k))$ embed $X$ homeomorphically into $[0,1{]}^{\mathbb{N}}$. Injectivity and inverse continuity follow by choosing $x_k$ close to a specified point and using the triangle inequality. The closure of the image gives $\tilde{X}$.

To show separability of $P(X)$, use finite sums of Dirac measures on a countable dense subset of $X$ with nonnegative rational weights summing to one. This is a countable family. Given $\mu$, cover a large-mass compact set by finitely many small balls centered in that subset, move the mass in each piece to its center, and move the remaining small mass to one fixed center. The bounded-Lipschitz error is at most the ball radius plus twice the exceptional mass. Approximate the resulting weights by rational weights. Thus these atomic measures are beta-dense.

Finally use a complete compatible metric $d$ on the Polish space $X$, as required for the granted total-boundedness-to-tightness assertion. A beta-Cauchy sequence is beta-totally bounded, and so is its closure in $P(X)$. By the allowed assertion this closure is uniformly tight; it is weakly closed because the weak and beta topologies agree. The compactness result just proved gives a convergent subsequence, and the Cauchy property forces the whole sequence to converge in beta. Consequently beta is complete for this choice of $d$, andIf the originally displayed compatible metric is incomplete, replace it by a complete compatible metric for this last argument; the weak topology itself is unchanged.

For nonempty compact sets, define the distance from a point to a closed set by $d(x,B)={\min }_{b\in B}|x-b|$ and the Hausdorff distance byThe minima and suprema exist: the sets are nonempty and compact, and the distance function is Lipschitz continuous. The Hausdorff distance is finite, nonnegative and symmetric. If it is zero, every point of $A$ belongs to the closed set $B$ and conversely, so $A=B$; the converse is immediate.

For $a\in A$, choose $b\in B$ attaining $d(a,B)$ and then $c\in C$ attaining $d(b,C)$. The triangle inequality gives $d(a,C)\le d(a,B)+d_H(B,C)$. Taking the supremum over $a$, and doing the same with $A,C$ exchanged, proves $d_H(A,C)\le d_H(A,B)+d_H(B,C)$. Hence this is a metric.

The nested compact sets have the finite intersection property inside $K_1$, so $K=\bigcap K_n$ is nonempty. It is a closed subset of $K_1$ and therefore compact. Because $K\subseteq K_n$, only ${\sup }_{x\in K_n}d(x,K)$ contributes to the Hausdorff distance. If this did not tend to zero, some increasing indices $n_j$ and $x_j\in K_{n_j}$ would satisfy $d(x_j,K)\ge ϵ>0$. A convergent subsequence in $K_1$ has limit $x$. For each fixed $m$, its tail belongs to the closed set $K_m$, so $x\in K$. But then $d(x_j,K)\le |x_j-x|\to 0$, a contradiction. Thus $\boxed{d_H(K_n,K)\to 0}$.

For $q\in \mathbb{Q}\setminus \{0\}$ write $q=p^{k}u/v$, where $k\in \mathbb{Z}$ and neither integer $u$ nor integer $v$ is divisible by $p$. Define the P-adic valuation $v_p(q)=k$ and the p-adic absolute value $|q{|}_p=p^{-k}$, with $|0{|}_p=0$. The p-adic metric isFactoring out the smaller power of $p$ from two rational numbers shows $v_p(q+r)\ge \min \{v_p(q),v_p(r)\}$, with the usual convention $v_p(0)=+\infty$. Consequently the ultrametric inequality givesTogether with symmetry, positivity and separation of points, this verifies that $d_p$ is a metric.

For the given geometric partial sums, use the finite geometric series identity. Since $p$ does not divide $p-1$,Thus the limit is the rational number $\boxed{1/(1-p)}$, even though the partial sums diverge in the ordinary real metric.

If $|a{|}_p>|b{|}_p$, the ultrametric inequality gives $|a+b{|}_p\le |a{|}_p$. Applying it to $a=(a+b)-b$ gives $|a{|}_p\le \max \{|a+b{|}_p,|b{|}_p\}$. Since $|b{|}_p<|a{|}_p$, this forces $|a+b{|}_p=|a{|}_p$. Interchanging $a,b$ handles the other unequal case, provingFinally, let $b$ be outside the open ball $B(a,\delta )$, so $d_p(a,b)\ge \delta$. If $d_p(b,c)<\delta$, the preceding unequal-norm identity gives $d_p(a,c)=d_p(a,b)\ge \delta$. Therefore $B(b,\delta )$ also lies outside $B(a,\delta )$. Its complement is open, so every p-adic open ball is also closed, that is, a clopen set.

Take $X$ to be the set $\mathbb{Q}$, equipped with the discrete metricEvery Cauchy sequence is eventually constant: its terms must eventually have mutual distance in this metric less than $1/2$. Consequently $X$ is a complete metric space. Its topology is the discrete topology, since every singleton is a ball of radius $1/2$ and every subset is a union of such balls.

For the codomain use the usual subspace topology on $\mathbb{Q}$. The inverse image of any open set there is a subset of the discrete $X$, hence is open. Therefore $f$ is continuous and plainly surjective. A continuous image of a complete metric space need not be complete. The two occurrences of $\mathbb{Q}$ have different domain and codomain metrics.

For a proper convex function $E$ on a real Hilbert space, its subdifferential at $y$ isIn a Banach space, interpret the pairing with an element of the dual space. For a chosen subgradient $p\in \partial E(y)$, the Bregman distance isIf also $q\in \partial E(x)$, its symmetric Bregman distance isThe subgradient inequality proves nonnegativity. These quantities depend on the chosen subgradients and generally are not metrics: they need not distinguish different points or obey the triangle inequality.

Let $x_n=(\pi /2+2\pi n{)}^{-1}$. The points $p_n=(x_n,1)$ lie on the specified graph and converge in ${\mathbb{R}}^2$ to $(0,1)$. They form a Cauchy sequence in the induced metric, but their only possible limit is absent from the space. Adding $(0,0)$ does not supply that missing limit. Hence the specified space is not complete.

For the unheaded request on $\mathbb{R}$, use the arctangent pullback metricPositivity, symmetry and the triangle inequality follow from the ordinary distance, and injectivity of $\arctan$ gives $d(x,y)=0$ only if $x=y$. The sequence $n$ is Cauchy for this metric, because $\arctan n\to \pi /2$, but cannot converge to any finite real $x$: that would require $\arctan x=\pi /2$. Thus this metric makes $\mathbb{R}$ incomplete, even though it induces its usual topology.