Write $T_n=h(\{x\in \mathbb{R}:|x|>n\})$, making explicit that these are subsets of $Y$. If $p\notin h[\mathbb{R}]$ and $U$ is an open neighbourhood of $p$, then $h([-n,n])$ is compact and hence closed in $Y$. The open set $U\setminus h([-n,n])$ still contains $p$ and meets the dense image of $h$. Such an intersection point belongs to $T_n$, so $p\in \overline{T_n}$ for every $n$.

Conversely, if $p=h(x)$ choose an integer $n>|x|$. Since $h$ is an embedding, there is an open neighbourhood $U$ of $p$ in $Y$ whose intersection with $h[\mathbb{R}]$ is $h((-n,n))$. It misses $T_n$, so $p\notin \overline{T_n}$. HenceFor the two ends of a real-line compactification, let $A_n^{+}=\overline{h((n,\infty ))}$ and $A_n^{-}=\overline{h((-\infty ,-n))}$. Each is nonempty compact and connected, being the closure of a connected image. The two nested families therefore have nonempty connected intersections $C^{+}$ and $C^{-}$. Since closure commutes with finite unions, $\overline{T_n}=A_n^{+}\cup A_n^{-}$. MoreoverFor if a point fails both intersections, it fails one positive member and one negative member, hence both members at the larger index by nesting.

If the remainder were a finite three-point set, its Hausdorff subspace topology would be discrete. Each nonempty connected subset $C^{\pm }$ would then be a singleton, so their union would have at most two points. This contradiction proves there is no three-point Hausdorff compactification of the real line. The argument does not assert that every possible remainder is finite.

A Hausdorff compactification of $X$ is an embedding $h:X\to Y$ into a compact Hausdorff space with dense image. The complement $Y\setminus h[X]$ is the remainder; a one-point or two-point compactification has a remainder of that cardinality.

For one point, take the unit circle andThe angle runs over $(-\pi ,\pi )$, so this is a homeomorphism onto the circle with $(-1,0)$ removed. That image is dense, and the circle is compact Hausdorff. For two points, take $Y=[-\pi /2,\pi /2]$ and $h_2(x)=\arctan x$. This is a homeomorphism onto its dense open interval, leaving exactly the two endpoints. Therefore both required compactifications exist; in the first the positive and negative ends meet, whereas in the second they remain distinct.

For a connected intersection of a closed filter base, the filter-base property implies the finite intersection property. Compactness, in its closed-set formulation, therefore gives $K=\bigcap \mathcal{F}\ne \varnothing$. It is closed in $X$, hence compact.

Suppose additionally that each base member is connected but $K$ is disconnected. Write $K=A\cup B$ with $A,B$ disjoint nonempty closed subsets of $K$. They are compact, hence closed in the Hausdorff space $X$. Normality of a compact Hausdorff space gives disjoint open sets $U,V$ containing $A,B$, respectively. The compact set $W=X\setminus (U\cup V)$ misses $K$. For every $w\in W$ some $F_w\in \mathcal{F}$ misses $w$, so the open sets $X\setminus F_w$ cover $W$. A finite subcover gives $F_1\cap \cdots \cap F_m\subseteq U\cup V$. Choose $F_0\in \mathcal{F}$ inside this finite intersection. If $W$ is empty, choose any base member instead.

Since $F_0$ contains $K$, it meets both $U$ and $V$; but it lies in their disjoint union. These two relatively open nonempty pieces disconnect $F_0$, a contradiction. Thus the intersection is nonempty and connected. Compactness is used to pass from an infinite exclusion condition to a single base member.

A filter base on a set $X$ is a nonempty family $\mathcal{F}$ of nonempty subsets such that for every $F,G\in \mathcal{F}$ there is $H\in \mathcal{F}$ with $H\subseteq F\cap G$. Repeated application gives a member contained in any specified finite intersection. In particular every finite intersection is nonempty, so the family has the finite intersection property. Its upward closure $\{A\subseteq X:\exists F\in \mathcal{F},F\subseteq A\}$ is the corresponding proper filter.

Take the natural numbers as the dense discrete subspace of their Stone-Čech compactification $\beta \mathbb{N}$. The ambient compact Hausdorff space is normal, and the Stone-Čech extension property makes this subspace $C^b$-embedded. It is not $C$-embedded: the continuous function $n\mapsto n$ is unbounded, whereas every continuous real-valued function on a compact space is bounded. The subspace is not closed, so this does not contradict the preceding closed-subspace theorem.

Let $h:Y\to \mathbb{R}$ be continuous, without a boundedness assumption. The homeomorphism $\theta (t)=t/(1+|t|)$ maps the real line onto $(-1,1)$, with inverse ${\theta }^{-1}(s)=s/(1-|s|)$. The bounded function $g=\theta \circ h$ extends by the preceding bounded construction to a continuous real-valued $G$ on $X$. Since $|G|<1$ on $Y$, the strict-range modification gives a continuous $\tilde{G}$ with range in $(-1,1)$ and the same restriction to $Y$. Thereforeis continuous everywhere: its denominator is pointwise positive. On $Y$ it equals ${\theta }^{-1}(\theta (h))=h$. Thus $Y$ is $C$-embedded. It is important to modify the extension before applying the inverse: an extension allowed to attain $1$ or $-1$ outside $Y$ would make this formula undefined there.

For a strict-range modification of a continuous extension, the closed sets $Y$ and $Z=\{x:|f(x)|\ge 1\}$ are disjoint. Choose by Urysohn's lemma a continuous $u:X\to [0,1]$ with $u{|}_Y=0$ and $u{|}_Z=1$, and defineThis function is continuous and agrees with $f$ on $Y$. On $Z$ it vanishes. Off $Z$ we have $|f|<1$ and $0\le 1-u\le 1$, so $|\tilde{f}|\le |f|<1$. Hence its range lies strictly inside $(-1,1)$ on all of $X$. The construction also covers empty $Y$ or empty $Z$ by a constant separating function where appropriate.

A C-embedded subset $Y\subseteq X$ is one when every continuous real-valued function on $Y$ extends continuously to $X$. It is a Cb-embedded subset when this is required for every bounded continuous real-valued function, with a bounded extension. These definitions concern the subspace topology on $Y$.

Here is the bounded Tietze extension construction from Urysohn's lemma. Let $g:Y\to \mathbb{R}$ be continuous with $|g|\le M$. The case $M=0$ has the zero extension. For $M>0$, the sets $A=\{g\le -M/3\}$ and $B=\{g\ge M/3\}$ are disjoint and closed in $X$, since $Y$ is closed. Urysohn's lemma gives a continuous $u_0:X\to [-M/3,M/3]$ taking the endpoint values on $A,B$. On $Y$,Indeed on $A$ or $B$ this follows from the prescribed endpoint; on their complement both $|g|$ and $|u_0|$ are at most $M/3$.

Repeat on the residual. With $M_n=M(2/3{)}^n$, choose continuous $u_n$ satisfying $\Vert u_n{\Vert }_{\infty }\le M_n/3$ and reducing the residual on $Y$ to at most $M_{n+1}$. The series $F={\sum }_{n\ge 0}{u}_n$ converges uniformly by the geometric majorant. Its sum is continuous, has $|F|\le \sum M_n/3=M$, and restricts to $g$, since the residual tends uniformly to zero. Thus every bounded continuous function extends, with its bound preserved, proving $C^b$-embeddedness without assuming the extension theorem being proved.

Urysohn's lemma states that if $X$ is normal and $A,B$ are disjoint closed subsets, there is a continuous function $u:X\to [0,1]$ with $u{|}_A=0$ and $u{|}_B=1$. The values on the rest of $X$ remain in that interval. Interchanging $A,B$ or composing with an affine function gives any two prescribed real endpoint values. This is functional separation of closed sets, stronger than merely choosing disjoint open neighbourhoods.

If the origin had a countable local basis $U_1,U_2,\dots$, replace it by the decreasing local basis $V_n={\bigcap }_{j\le n}{U}_j$. Each $V_n$ meets the dense set $Y$, so choose $p_n\in V_n\cap Y$. Given any neighbourhood $U$ of the origin, some $U_j$ is contained in $U$; for every $n\ge j$ we then have $p_n\in V_n\subseteq U_j\subseteq U$. Thus $p_n$ converges to the origin, contradicting the preceding part. Consequently first countability fails already at the origin, and $X$ is not first countable.

Suppose for contradiction that $p_n=(u_n,v_n)\in Y$ converges to the origin in the initial topology. The coordinate projections then give $u_n\to 0$ and $v_n\to 0$ in the usual real line. Rectangular bumps detecting nonconvergence, using anisotropic supports, avoids any unsupported regularity assertion about a radial maximum. SetThe radii are positive since neither coordinate of $p_n$ vanishes. Each rectangular support misses both coordinate axes, so $B$ vanishes on those axes. For a fixed $y=c\ne 0$, all but finitely many vertical support intervals miss $c$, because their centres and radii tend to zero. Hence $B(x,c)$ is a finite maximum of continuous functions of $x$, and is continuous. At $c=0$ it is identically zero. The same argument with the coordinates exchanged proves separate continuity everywhere. Thus $B$ is continuous for the initial topology.

However $B(p_n)=1$ for every $n$, whereas $B(0,0)=0$. This contradicts preservation of convergent sequences by a continuous function. Therefore no sequence from $Y$ converges to the origin.

The radial hint in the scan is not separately continuous for every candidate sequence: for $p_n=(2^{-n},2^{-2n})$, its value at the origin is zero, while its value at $(2^{-n},0)$ is at least $1-2(2^{-2n})/\sqrt{2^{-2n}+2^{-4n}}$, which tends to one. The rectangular construction supplies the needed valid proof independently of that hint.

For a fixed nonzero second coordinate $c$, the rational expression defining $h(x,c)$ is continuous on the half-line where $xc>0$. On the opposite half-line it is constant. At $x=0$ the ratio tends to $1$, agreeing with the value on the other side. If $c=0$ the entire restriction is constant. By symmetry the restrictions with the first coordinate fixed are continuous too. Hence $h$ is separately continuous, and so is $h_t(x,y)=h(x-t,y-t)$ for every real $t$.

On the diagonal $D$, $h_t(t,t)=1$, while $h_t(s,s)=0$ for $s\ne t$. ThereforeEach diagonal singleton is open in the induced topology; arbitrary unions show that $D$ has the discrete topology.

If $X$ were metrizable, its countable dense subset from the preceding part would make it second countable. Indeed balls with rational radii about dense points contain a countable basis after the usual smaller-ball argument. Every subspace of a second-countable space is second countable. A discrete second-countable space is countable, since each singleton must itself occur as a basis member. But $D$ is uncountable. This contradiction proves $X$ is not metrizable.

Fix a horizontal line $L=\mathbb{R}\times \{c\}$. The restriction of every member of $\mathcal{H}$ to $L$ is continuous in its usual topology. Therefore every subbasic open set intersects $L$ in a usual open set, so the induced topology is no finer than the usual one. Conversely the first coordinate projection is continuous for the initial topology, and its inverse images induce every usual open set on $L$. The two topologies on $L$ are equal. The second projection gives the identical argument for a vertical line.

Write $F={\overline{Y}}^X$. For every nonzero rational $r$, the intersection of $Y$ with $\{r\}\times \mathbb{R}$ is dense in that vertical line, since the nonzero rational numbers are dense in the real line. Hence $\{r\}\times \mathbb{R}\subseteq F$. For each real $c$, it follows that $F\cap (\mathbb{R}\times \{c\})$ contains $(\mathbb{Q}\setminus \{0\})\times \{c\}$ and is closed in that line. This dense subset forces the whole horizontal line into $F$. Thus $F=X$: $Y$ is a countable dense subset. The two successive uses of line density are necessary; the initial topology is finer than the Euclidean topology, so Euclidean density alone would not suffice.

Let $\mathcal{H}$ be the set of separately continuous real-valued functions on the plane. Take as a subbasis all sets $h^{-1}(U)$, where $h\in \mathcal{H}$ and $U$ is open in the usual real line. Unions of finite intersections of these sets form a topology, and every $h$ is continuous for it by construction. Conversely every topology making every $h$ continuous must contain the subbasis, hence this entire generated topology. Thus the required coarsest topology exists and is unique: it is the initial topology of the family $\mathcal{H}$.

The family is a set, being a subset of ${\mathbb{R}}^{{\mathbb{R}}^2}$, so this construction has no proper-class issue. The coordinate projections belong to $\mathcal{H}$; consequently this topology contains the ordinary Euclidean topology. This last observation will also make convergence in the new topology imply ordinary coordinatewise convergence.

Every constructible set is well-orderable: the constructible hierarchy has a canonical well-order, obtained by ordering the stage of first appearance, logical formula codes and the already ordered finite parameter lists. Its restriction to any set belonging to $L$ is a well-order in $L$ and remains a well-order in the ambient universe. This does not imply that arbitrary ambient sets are well-orderable.

If $a\in L$ is finite, all its subsets are finite combinations of its elements and hence constructible. Moreover the finite set of all those subsets is constructible by the finite set operations available in $L$. Thus the full ambient power set $\mathcal{P}(a)$ belongs to $L$.

Conversely, suppose $a\in L$ is infinite and that its full ambient power set belonged to $L$. A well-order of $a$ supplies an injection $j:\omega \to a$ by taking its first countably many elements. The mapis injective. But $\mathcal{P}(a)\in L$ would make that set well-orderable in the ambient universe, and pulling its well-order back along this injection would well-order $\mathcal{P}(\omega )$. This contradicts the hypothesis. HenceThe superscript emphasizes that the source's final claim concerns the ambient power set. The internal power set ${\mathcal{P}}^L(a)$ exists in $L$ for every $a\in L$ by part (ii), including infinite ones.

Fix $a\in L$. In the ambient universe form the setby power set and separation, using definability of $L$. This is the collection which must serve as the power set of $a$ inside $L$; it need not be the entire ambient power set.

For each $y\in S$, let ${\alpha }_y$ be the least ordinal with $y\in L_{{\alpha }_y}$. Ambient replacement is used to collect the ordinals ${\alpha }_y$ into a set. Their supremum, enlarged if necessary to include a stage containing $a$, gives an ordinal $\beta$ such that $a\in L_{\beta }$ and every $y\in S$ belongs to $L_{\beta }$.

Because $L_{\beta }$ is transitive and $a\in L_{\beta }$, the bounded property $y\subseteq a$ for $y\in L_{\beta }$ has its actual meaning when evaluated in that structure: every element of $y$ already lies in $L_{\beta }$, and membership is unchanged. It follows thatThis is a definable subset of $L_{\beta }$ using the parameter $a$, so $S\in \mathrm{Def}(L_{\beta })=L_{\beta +1}\subseteq L$. Every member of $L$ which is a subset of $a$ is in $S$, and every member of $S$ is such a subset. Thereforewhich is exactly the power-set axiom in $⟨L,\in ⟩$. The replacement step is indispensable to this proof: definability over one stage requires first bounding all the constructible subsets by one common stage.

For a set $M$, let $\mathrm{Def}(M)$ consist of the subsets of $M$ definable over the structure $⟨M,\in ⟩$ by a first-order formula with finitely many parameters from $M$. It is a set contained in $\mathcal{P}(M)$: satisfaction for a set structure and the set of logical formula codes give a uniform set construction of these definable subsets. If $M$ is transitive, every $a\in M$ is one such subset, defined over $M$ by $z\in a$, so $M\subseteq \mathrm{Def}(M)$.

Define the constructible hierarchy by transfinite recursion:Gödel's constructible universe is the definable class in set theoryAt successor stages only subsets definable over the previous stage are admitted, in contrast to the full power-set construction of the cumulative hierarchy. The stages are transitive and increasing, as permitted in the question, and their union contains every ordinal. Finite parameters are essential in the definition of $\mathrm{Def}$.

In the usual arithmetic under the axiom of choice of cardinal numbers, take the singular strong-limit cardinalCantor's theorem makes this sequence strictly increasing. It is countable and cofinal in its supremum, so $\mathrm{cf}(\kappa )\le {\aleph }_0$. No finite subset is cofinal in this limit ordinal, hence $\mathrm{cf}(\kappa )={\aleph }_0$. ThereforeThis construction needs no assumption such as the Continuum hypothesis. The conventional use of cardinals here includes well-orderability of these power sets, supplied by the axiom of choice. It is a separate setting from the explicit failure of well-orderability hypothesized in question 8; one must not impose choice on that later hypothesis.

For an infinite cardinal regarded as an ordinal, its cofinality $\theta =\mathrm{cf}(\kappa )$ is the least cardinality of a cofinal subset of $\kappa$. Equivalently there is a family $({\alpha }_i{)}_{i<\theta }$ of ordinals below $\kappa$ whose supremum is $\kappa$, and no smaller indexed family has that property. Since an infinite initial ordinal is a limit ordinal, $\theta$ is infinite and in particular nonzero.

Use the diagonal mechanism of König theorem for cardinal numbers directly. Suppose $F:\kappa \to {}^{\theta }\kappa$ were surjective. For each $i<\theta$, the sethas cardinality at most $|{\alpha }_i|<\kappa$. Choose the least $b_i\in \kappa \setminus E_i$. The resulting function $b:\theta \to \kappa$ is not in the range: for any $\alpha <\kappa$, cofinality supplies $i$ with $\alpha <{\alpha }_i$, and then $b_i\ne F(\alpha )(i)$.

Thus no such surjection exists. Constant functions provide an injection $\kappa \to {}^{\theta }\kappa$, and an injection in the reverse direction would give a surjection by the same inverse-and-default construction as in part (i). ConsequentlyThis argument also handles $\kappa ={\aleph }_0$, when cofinal initial segments are finite; it does not incorrectly assume that smaller cardinals themselves must form a cofinal sequence in every infinite cardinal.

König theorem for cardinal numbers states that if ${\kappa }_i<{\lambda }_i$ for each index $i\in I$, thenRepresent each cardinal by its initial ordinal, let $A=\{(i,\alpha ):i\in I,\alpha <{\kappa }_i\}$, and let $B={\prod }_{i\in I}{\lambda }_i$.

First there is an injection $A\to B$: encode $(i,\alpha )$ by the function which is zero away from $i$ and takes value $\alpha +1$ at $i$. The nonzero coordinate identifies $i$, and its value identifies $\alpha$. It is valid because $\alpha +1\le {\kappa }_i<{\lambda }_i$; an index with ${\kappa }_i=0$ contributes no domain points.

For strictness suppose a surjection $F:A\to B$ existed. At coordinate $i$, the values $F(i,\alpha )(i)$ for $\alpha <{\kappa }_i$ occupy fewer than ${\lambda }_i$ possibilities. Choose $b_i$ as the least ordinal below ${\lambda }_i$ absent from that set. The resulting $b=(b_i)\in B$ differs from $F(i,\alpha )$ at coordinate $i$ for every $(i,\alpha )\in A$, contradicting surjectivity.

An injection $B\to A$ would produce a surjection $A\to B$ by inverting on its image and assigning the zero function elsewhere. It is therefore impossible. The injection in the first direction and impossibility of the reverse give the strict cardinal inequality. This diagonal argument uses the canonical well-orders of the specified cardinals to choose missing coordinate values, rather than an unexplained arbitrary choice at that step.