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Let $X=\mathbb{R}$ with the usual metric and let $Y=\mathbb{R}$ with the discrete metric, and take $f(x)=x$. Every subset of $Y$ is an open set, so the image of every open subset of $X$ is open: $f$ is an open map.

It is not continuous at any point. For $ϵ=1/2$, every $\delta >0$ allows a distinct $y$ with $|y-x|<\delta$, but the distance between $f(y)$ and $f(x)$ in the discrete metric is 1. Equivalently, the preimage of the open singleton $\{x\}\subset Y$ is not open in $X$. Thus $\boxed{\text{an open map need not be continuous}}$ even between metric spaces.

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.

A metric space metric $d:X\times X\to [0,\infty )$ satisfies positivity, $d(x,y)=0$ exactly when $x=y$, symmetry, and the triangle inequality. A subset $U\subseteq X$ is an open set when every $x\in U$ has some $\epsilon >0$ with the open ball $B_d(x,\epsilon )\subseteq U$.

Both $\varnothing$ and $X$ are open. An arbitrary union of open sets is open because a point lies in one member supplying a ball. A finite intersection is open because the minimum of the finitely many available radii supplies a ball. Thus the metric-open sets satisfy the topology axioms.

For $C[0,1]$, the Cauchy-Schwarz inequality gives $d_1(f,g)\le d_2(f,g)$, so the $d_2$ topology is at least as fine as the $d_1$ topology. To see strictness, define the continuous triangular spikesThenThus $f_n$ converges to zero in the L1 norm but not in the L2 norm. The two metrics induce different topologies.

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 ultrafilter space. An ultrafilter $U$ on $\mathbb{N}$ is a proper filter on a set: it is upward closed, closed under finite intersections, excludes $\varnothing$, and contains exactly one of $A,A^c$ for each $A\subseteq \mathbb{N}$. The Stone-Čech compactification of the natural numbers is the set of these ultrafilters, with basic setsThey form a basis since $\hat{A}\cap \hat{B}=\widehat{A\cap B}$ and $\hat{\mathbb{N}}=\beta \mathbb{N}$. Moreover $(\hat{A}{)}^c=\widehat{A^c}$, so each is a clopen set. The principal ultrafilter at $n$ embeds $\mathbb{N}$ in this space; every nonempty $\hat{A}$ contains such a point, so the embedded copy is dense.

Distinct ultrafilters $U,V$ have some $A\in U$ with $A\notin V$. Then $A^c\in V$, and $\hat{A},\widehat{A^c}$ are disjoint open sets separating them. Thus the space is Hausdorff.

For compactness, any open set cover can be refined to a cover by basic sets. If $\{\widehat{A_i}\}$ has no finite subcover, the sets $A_i^c$ have the finite intersection property: otherwise finitely many $A_i$ cover $\mathbb{N}$, and their basic sets cover every ultrafilter. Extend the proper filter on a set generated by these complements to an ultrafilter $U$. This extension follows from Zorn lemma: unions of chains of proper filters remain proper, and a maximal proper filter decides every $A$ versus $A^c$. Indeed, if adding $A$ made the filter improper, some existing filter member would be disjoint from $A$, forcing $A^c$ into the filter. This $U$ contains every $A_i^c$ and misses every $\widehat{A_i}$, a contradiction. HenceThis proves the compact Hausdorff topology on ultrafilters directly.

An idempotent. The addition on the Stone-Čech compactification of the natural numbers can be defined byAssociativity is supplied. For fixed $V$, the inverse image of $\hat{A}$ under $U\mapsto U+V$ is $\widehat{\{n:A-n\in V\}}$, so this translation is continuous directly from the definition. This makes a left-topological semigroup in the convention used here.

We prove the Ellis–Numakura lemma in this case. Among nonempty compact subsemigroups choose one, $K$, minimal by inclusion. It exists by Zorn lemma: a descending chain has nonempty intersection by compactness, and that intersection is again a compact subsemigroup. For $a\in K$, the set $K+a$ is a nonempty compact subsemigroup contained in $K$. Compactness follows from continuity; closure under addition follows fromMinimality gives $K+a=K$. Hence there is $b\in K$ with $b+a=a$. The setis nonempty, compact, and a subsemigroup: if $x,y\in L$, then $(x+y)+a=x+(y+a)=x+a=a$. Thus minimality gives $L=K$, and in particular $a+a=a$. ThereforeThis is an idempotent ultrafilter on the natural numbers.

Isolating the sequence terms. Now use the hypothetical convergent sequence of distinct ultrafilters. For each $n$, choose $C_n\in U_n$ with $C_n^c\in U$, possible since $U_n\ne U$. Convergence implies $C_n^c\in U_m$ for all sufficiently large $m$. Thus only finitely many indices $m\ne n$ have $C_n\in U_m$. Call this finite set $E_n$.

For each $m\in E_n$, choose $D_{n,m}\in U_n$ with $D_{n,m}\notin U_m$, using distinctness. SetFinite-intersection closure puts $A_n$ in $U_n$. For $m\ne n$, either $C_n\notin U_m$ or one of the chosen $D_{n,m}$ is absent from $U_m$; upward closure then prevents $A_n$ from belonging to $U_m$. Hence

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

The answer is no. Let $X_0=\mathbb{N}$, regarded as one point of the space of infinite subsets of the natural numbers, and putThe displayed union makes $O$ an open set in the Ellentuck topology. On the other hand, every basic neighbourhood $[s,A]$ of $X_0$ contains a different infinite set: keep $s$ and remove one element from its infinite remaining tail. This other point belongs to $O$. Therefore $X_0$ is in the closure of $O$ but not in $O$, soThe fact that basic neighbourhoods are clopen sets does not make arbitrary unions of them closed.

The two Ramsey properties and the topologies. Write $[\mathbb{N}{]}^{\omega }$ for the space of infinite subsets of the natural numbers, and $[A{]}^{\omega }$ for the infinite subsets of $A$. The positive integers are used throughout. A Ramsey set of infinite subsets $E$ has the property that every infinite $A$ contains an infinite $B$ for which $[B{]}^{\omega }$ lies wholly in $E$ or wholly in $E^c$.

For finite $s$ and infinite $A$ above $\max s$, letThese are the basic neighbourhoods of the Ellentuck topology, denoted $\ast$ in the question. They form a basis: if $X$ belongs to two such sets, their stems are comparable initial segments of $X$; take the longer stem and the intersection of the two tails beyond that stem. This intersection contains the infinite remaining tail of $X$ and yields a basic set contained in both.

A completely Ramsey set $E$ allows such a homogeneous refinement $[s,B]$ inside every $[s,A]$, with $B\subseteq A$ infinite and the stem $s$ kept fixed. Taking $s=\varnothing$ shows that complete Ramseyness implies Ramseyness. The ordinary topology on infinite subsets $\tau$ has basic cylindersIt is the usual product topology on increasing enumerations and is coarser than the Ellentuck topology.

Examples distinguishing the properties. A non-Ramsey example uses the axiom of choice. Let $\kappa =2^{{\aleph }_0}$ and enumerate all infinite sets as $(A_{\alpha }{)}_{\alpha <\kappa }$. Each $[A_{\alpha }{]}^{\omega }$ has cardinality $\kappa$: an enumeration of $A_{\alpha }$ gives the upper bound, and choosing one element from each of infinitely many disjoint pairs gives an injection from all binary sequences for the lower bound.

By transfinite recursion, choose two distinct, previously unused elements $X_{\alpha },Y_{\alpha }\in [A_{\alpha }{]}^{\omega }$ at each stage. Fewer than $\kappa$ elements have been used at stage $\alpha <\kappa$, so this is possible without any assumption on the Continuum hypothesis. Put $R=\{X_{\alpha }:\alpha <\kappa \}$. All the $Y_{\alpha }$ remain outside $R$, since subsequent choices also avoid previously chosen elements. Every $[A{]}^{\omega }$ contains one $X_{\alpha }$ and one $Y_{\alpha }$, so $R$ is not a Ramsey set of infinite subsets. This is the non-Ramsey set from transfinite selection construction.

Perform the same construction on $T=\mathbb{N}\setminus \{1\}$, giving a non-Ramsey set $R_T\subseteq [T{]}^{\omega }$, and putFor any infinite $A$, the infinite set $B=A\setminus \{1\}$ has $[B{]}^{\omega }\cap E=\varnothing$, so $E$ is a Ramsey set of infinite subsets. But no $[\{1\},B]$ with infinite $B\subseteq T$ is homogeneous, because $[B{]}^{\omega }$ meets both $R_T$ and its complement. Therefore $E$ is not a completely Ramsey set.

Every star-open set is completely Ramsey. We give the full fusion proof for open Ellentuck sets, without quoting an infinite Ramsey theorem. Fix a star-open set $O$ and a basic set $[s,A]$. For finite $a\subseteq A$ and an infinite tail $D\subseteq A$ above $\max a$, say $D$ accepts $a$ if $[s\cup a,D]\subseteq O$; it rejects $a$ if no infinite subset of $D$ accepts $a$. Both properties persist under passing to infinite subsets. Every infinite tail can be thinned to decide $a$: take an accepting subset if one exists, and otherwise the original tail rejects it.

First thin $A$ to decide $a=\varnothing$. If it accepts, the required homogeneous neighborhood has already been found. Otherwise keep a rejecting tail. We now fuse to an infinite set $B$ deciding every finite $a\subseteq B$, where the decision for $a$ is made by $B/a=\{b\in B:b>\max a\}$.

To do this, choose $b_1<b_2<\cdots$ successively from an infinite current tail. After choosing $b_j$, thin the tail above $b_j$ to decide each of the finitely many subsets of $\{b_1,\dots ,b_j\}$, one after another, before choosing the next element. For any finite $a\subseteq B$, at the stage when its largest element is selected, the current tail decides $a$ and contains every later selected element. Heredity therefore gives a decision on $B/a$. For $a=\varnothing$, rejection persists from the initial tail.

If $B/a$ rejects $a$, only finitely many $n\in B/a$ can have $B/n$ accepting $a\cup \{n\}$. Otherwise let $D$ be the infinite set of those $n$. Every member of $[s\cup a,D]$ has a first new element $n$, and its remaining tail lies in $B/n$. It therefore belongs to $O$ by acceptance of $a\cup \{n\}$. This would make $D$ accept $a$, contradicting rejection.

Thin $B$ a second time to $C=\{c_1<c_2<\cdots \}$ so that every finite $a\subseteq C$ is rejected. Start with the rejected empty set. At a finite stage, all subsets of the selected elements are rejected. For each of these finitely many subsets, avoid its finitely many next elements that would give an accepted extension, and choose the next $c_j$ beyond all excluded elements. Since $B$ decides every finite extension, each new extension is rejected. Induction gives the asserted property of $C$.

We claim $[s,C]\cap O=\varnothing$. If $X\in [s,C]\cap O$, openness provides a basic set $[t,D]\subseteq O$ containing $X$. Extend $t$, if necessary, to an initial segment $u$ of $X$ containing $s$. Write $u=s\cup a$ with finite $a\subseteq C$. The infinite tail $X\setminus u$ lies in $B/a$ andIt is therefore an accepting subtail for $a$, contradicting rejection. We have found either an accepting neighborhood or a neighborhood disjoint from $O$. Thus

Basic neighborhoods are clopen. Consider $[s,A]$ and a point $X$ outside it. If $s$ is not an initial segment of $X$, take an initial segment $t$ of $X$ of length at least $|s|$; its ordinary cylinder $[t]$ cannot meet $[s,A]$. Otherwise some element of $X\setminus s$ is outside $A$; take $t$ through that element, and again $[t]$ is disjoint from $[s,A]$.

Thus the complement of $[s,A]$ is already open in the ordinary topology on infinite subsets, and hence in the finer Ellentuck topology. A basic neighbourhood is open by definition, so

Use the Fourier transform convention $\hat{f}(\xi )=\int e^{-ix\cdot \xi }f(x)dx$ and $D_j=-i{\partial }_{x_j}$. Thus $D_j$ has Fourier symbol ${\xi }_j$. This fixes the factors of $i$ in the formula for a differentiated Dirac delta distribution below.

With the Japanese bracket $⟨\theta ⟩=(1+|\theta {|}^2{)}^{1/2}$, the symbol class $\mathrm{Sym}(X,{\mathbb{R}}^k;N)=S_{1,0}^N(X\times {\mathbb{R}}^k)$ consists of smooth functions $a$ such that, for every compact set $K\subset X$ and all multi-indices $\alpha ,\beta$,Here $N$ is one fixed finite real order; the constants may depend on $K$. Differentiation in $x$ preserves the symbol class order, whereas differentiation in $\theta$ lowers it.

A phase function is real-valued and smooth on $X\times ({\mathbb{R}}^k\setminus \{0\})$, is a positively homogeneous function of degree one in $\theta$, and has nonzero total differential:The nonvanishing condition concerns both sets of variables, not just ${\nabla }_{\theta }\Phi$. The homogeneous convention is imposed away from $\theta =0$; a smooth completion at low frequency is another equivalent convention for the high-frequency construction. Low-frequency changes contribute a smooth function of $x$.

Choose a cutoff function $\rho (\theta )$ equal to one near zero. The low-frequency part with oscillatory integral amplitude $\rho a$ is an ordinary convergent integral and defines a smooth function: all $x$ derivatives of $\Phi$ are $O_K(|\theta |)$ near zero, so differentiation under this integral preserves integrability. For the remaining part define, when $\theta \ne 0$,Then $L{e}^{i\Phi }=e^{i\Phi }$. On $K\times \{\theta :|\theta |=1\}$, compactness and the phase function condition give a positive lower bound for $G$. Homogeneity consequently gives $G\ge c_K|\theta {|}^2$. The coefficient of each $x$ derivative in $L$ has symbol class order $-1$, and the coefficient of each $\theta$ derivative has order zero.

If $L=\sum b_j{\partial }_{x_j}+\sum c_{\ell }{\partial }_{{\theta }_{\ell }}$, its formal transpose isThis is the bilinear transpose for integration by parts, without complex conjugation. It lowers the symbol class order by one. For a test function $\phi$ supported in $K$, choose a nonnegative integer $r>N+k$ and setBoth integrals are absolutely convergent, because the last oscillatory integral amplitude has order $N-r<-k$ and compact support in $x$.

To verify that this defines the intended oscillatory integral, take any $\chi \in C_c^{\infty }({\mathbb{R}}^k)$ equal to one near zero and insert $\chi (\epsilon \theta )$ in the original integral. Repeated integration by parts gives the preceding expression with $(L^t{)}^r$ applied also to this cutoff function. Its derivatives satisfy uniform symbol class bounds: on the annulus where they are nonzero, ${\epsilon }^{|\beta |}\le C⟨\theta {⟩}^{-|\beta |}$. The transformed integrands are bounded by an integrable multiple of $⟨\theta {⟩}^{N-r}$. The dominated convergence theorem therefore provesIt also proves independence of the cutoff function, the chosen $r$, and the integration-by-parts representation.

At most $r$ derivatives fall on $\phi$, so the same estimates giveThis proves linearity and continuity on the space of test functions. It also proves the finite order of an oscillatory integral distribution: the derivative bound uses the same $r$ for every $K$, although $C_K$ changes.

For $x_0\in X$, take $k=n$, $\Phi =(x-x_0)\cdot \theta$, and $a=(2\pi {)}^{-n}{\theta }^{\alpha }$. This phase function is valid because ${\nabla }_x\Phi =\theta \ne 0$ for nonzero $\theta$, and the oscillatory integral amplitude is in symbol class $S_{1,0}^{|\alpha |}$. The required identity isIndeed, extending a test function by zero outside $X$ and integrating in $x$ first gives the absolutely convergent expressionHere Fourier inversion applies because $\phi$ is a Schwartz function. By the definition of a distributional derivative, the last expression is precisely $⟨D^{\alpha }{\delta }_{x_0},\phi ⟩$. If $D^{\alpha }$ is instead used for plain ${\partial }^{\alpha }$, the oscillatory integral amplitude in the boxed formula is $(2\pi {)}^{-n}(i\theta {)}^{\alpha }$, and the pairing is $(-1{)}^{|\alpha |}{\partial }^{\alpha }\phi (x_0)$.

Not every distribution is one oscillatory integral of the stated class. On $X=\mathbb{R}$, consider the locally finite distribution of unbounded orderOnly finitely many differentiated Dirac delta distributions contribute to each test function, so $T$ is a distribution. Near $j$ it is exactly $D^j{\delta }_j$, which cannot satisfy a bound using only $r<j$ derivatives. Explicitly choose $\eta \in C_c^{\infty }((-1/4,1/4))$ with ${\eta }^{(j)}(0)\ne 0$ and useThe derivatives through order $r$ remain bounded for $0<\epsilon \le 1$, whereasAny single oscillatory integral with finite $k$ and a fixed finite symbol class order $N$ has the uniform order bound just proved, so it cannot equal $T$. The same construction works on any nonempty open set in positive dimension by choosing points that leave every compact set and taking locally finite differentiated Dirac delta distributions there. This obstruction concerns a single fixed-order oscillatory integral, rather than local representations or an infinite sum of them.

A subset of $\mathbb{R}$ is a Borel set when it belongs to the Borel sigma-algebra, the smallest sigma-algebra containing every open set of $\mathbb{R}$.

Write $I^{+}(U)=U=I^{-}(U)$. Since every chronological future and chronological past is open, $U$ is an open set. We show its complement is also open.

Take $p\notin U$ and choose $p_{-}\ll p\ll p_{+}$ in a convex normal neighborhood. The diamond $V=I^{+}(p_{-})\cap I^{-}(p_{+})$ is a neighborhood of $p$. If $x\in V\cap U$, then $p_{-}\ll x$ and $U=I^{-}(U)$ imply $p_{-}\in U$; then $p_{-}\ll p$ and $U=I^{+}(U)$ imply $p\in U$, a contradiction. Thus $V$ lies in the complement of $U$.

Hence $U$ is nonempty, open, and closed. Since $\mathcal{M}$ is a connected space,

Assume inverse images of open sets are open. If $x_n\to a$ and $V$ is any open neighbourhood of $f(a)$, then $f^{-1}(V)$ is an open neighbourhood of $a$. By the definition of convergence in a metric space, $x_n\in f^{-1}(V)$ eventually, and hence $f(x_n)\in V$ eventually. Thus $f(x_n)\to f(a)$.

The identity theorem states that if two holomorphic functions on a connected open set $\mathcal{D}$ agree on a subset having an accumulation point in $\mathcal{D}$, then they agree throughout $\mathcal{D}$. In particular, agreement on any nonempty open subset is sufficient.

An $m$ by $n$ matrix has matrix rank $m$ exactly when at least one of its $m$ by $m$ minors has nonzero determinant. For each multi-index $I$, the setis open in the vector space of all real $m$ by $n$ matrices, because the determinant is continuous. Their union is $X_{m,n}$, so $X_{m,n}$ is itself open in ${\mathbb{R}}^{mn}$. It therefore inherits the standard smooth manifold structure and has dimension $mn$. This is the full-row-rank matrix manifold.

Suppose first that $X$ is connected and $f:X\to \mathbb{Z}$ is continuous. Its image is connected by the continuous image of a connected space, but $\mathbb{Z}$ is discrete, so its only connected nonempty subsets are singletons. Hence $f$ is constant.

Conversely, if $X=U\cup V$ is a disconnection into nonempty disjoint open sets, the function equal to zero on $U$ and one on $V$ is continuous and nonconstant. This proves the integer-valued function criterion for connectedness.

Now let $f:X\to \mathbb{Z}$ be continuous under the hypotheses on the family $\mathcal{A}$. Each restriction $f{|}_A$ is constant because $A$ is connected. If $A,B\in \mathcal{A}$, a point of $A\cap B$ shows that their two constants agree. Since the sets cover $X$, $f$ is constant on $X$, and the criterion proves that $X$ is connected. This is the pairwise-intersecting connected cover argument.

Finally, fix $y_0\in Y$. For each $x\in X$, the setis connected: its two connected pieces meet at $(x,y_0)$. The sets $A_x$ cover $X\times Y$ and any two share $X\times \{y_0\}$. The preceding result proves that $X\times Y$ is connected, giving the product of connected spaces result.

Differentiate the affine relation with respect to $x$ and $y$:Using the Cauchy-Riemann equations, the second equation becomesThusThe determinant is $-(a^2+b^2)\ne 0$, so $u_x=v_x=0$ and therefore $f^{\prime }=0$. A holomorphic function with zero derivative on a connected open set is constant. Hence $f$ is constant, as in analytic function with image in an affine real line.

A topological space is compact if every open cover has a finite subcover. It is Hausdorff if every pair of distinct points has disjoint open neighbourhoods.

Let $Y$ be closed in compact $X$, and let $\{U_i\}$ be an open cover of $Y$ by sets open in $X$. Adding the open set $X\setminus Y$ gives a cover of $X$, which has a finite subcover. Removing $X\setminus Y$ leaves a finite subcover of $Y$. Thus every closed subspace of a compact space is compact.

Now let $A,B$ be disjoint closed subsets of a compact Hausdorff space. They are compact. For each $a\in A$ and $b\in B$, choose disjoint open sets $U_{a,b}\ni a$ and $V_{a,b}\ni b$. Fixing $a$, finitely many $V_{a,b_j}$ cover $B$. PutThen $U_a$ contains $a$, $V_a$ contains $B$, and they are disjoint. Finitely many $U_{a_i}$ cover $A$. Thereforeare disjoint open neighbourhoods of $A$ and $B$. This proves the normality of a compact Hausdorff space.

Let $X$ be compact Hausdorff, let $x\in X$, and let $U$ be a neighbourhood of $x$. Choose an open $O$ with $x\in O\subseteq U$. Apply the separation result from part (a) to the disjoint closed sets $\{x\}$ and $X\setminus O$. There is an open $V\ni x$ whose closure lies in $O$. Thenis compact, contains the neighbourhood $V$ of $x$, and lies in $U$. Thus $X$ is locally compact.

Conversely, suppose $X$ is locally compact Hausdorff and $A\cap K$ is closed in every compact $K\subseteq X$. For $x\notin A$, choose a compact neighbourhood $K$ of $x$ and an open set $U$ withSince $A\cap K$ is closed in $K$, there is an open set $W\subseteq X$ such thatThen $U\cap W$ is an open neighbourhood of $x$ disjoint from $A$. Thus $X\setminus A$ is open andThis is the compactly detected closed-set theorem in a locally compact Hausdorff space.

The inequalities implyThus the two metrics induce the same open sets and are equivalent.

The converse fails. On $\mathbb{R}$, letThe metrics are equivalent because $\arctan :\mathbb{R}\to (-\pi /2,\pi /2)$ is a homeomorphism, butso no positive lower comparison constant exists. This is an equivalent metrics need not be bi-Lipschitz equivalent example.

Equivalent metrics on the codomain also need not give the same uniform convergence. Take $X=\mathbb{N}$, $Y=\mathbb{R}$, and use the two metrics above. DefineThenso $f_n\to f$ uniformly for $d_2$, whereasso convergence is not uniform for $d_1$. This is the equivalent codomain metrics need not preserve uniform convergence phenomenon.

A covering space is a map $p:\tilde{X}\to X$ such that every $x\in X$ has an open neighbourhood $U$ for whichand every restriction $p:U_{\alpha }\to U$ is a homeomorphism.

For path lifting, cover the compact image of $\gamma$ by evenly covered sets. The Lebesgue number lemma supplies a subdivisionfor which each $\gamma ([t_{j-1},t_j])$ lies in one such set. Starting in the sheet containing ${\tilde{x}}_0$, lift the first segment with the inverse of that sheet. Its endpoint selects the sheet for the next segment, and induction constructs a continuous lift. Two lifts starting at the same point agree on the first segment because the relevant sheet map is injective, and the same argument at successive endpoints proves uniqueness.

Write $q:Y\to Y/\varphi$ for the quotient map. The identity $p\circ Q=q\circ {\mathrm{pr}}_Y$ on $Y\times \mathbb{Z}$, where $Q$ is the quotient defining $\widetilde{Y/\varphi }$, proves continuity of $p$ by the quotient property. Away from the two gluing regions, a small open set lifts to one copy in every level. Near a glued point choose $U⋐A$ and use the matched pairAfter the prescribed identifications, each such pair maps homeomorphically to $q(U\cup \varphi (U))$, and the pairs are disjoint for different $i$. Thus $p$ is a covering map.

Each copy of $Y$ is path-connected, and adjacent copies meet through the identified copies of the nonempty set $A$, so $\widetilde{Y/\varphi }$ is path-connected. Translationis a deck transformation. Its powers act freely and transitively on every fibre, so the covering is regular. The covering-space subgroup and deck-group quotient then gives