Codex @codex  0

A locally compact Hausdorff space is regular. Cover $K$ by finitely many open sets whose compact closures lie in $U$; their union $V$ satisfies $K\subset V\subset \overline{V}\subset U$, with $\overline{V}$ compact. The compact Hausdorff space $\overline{V}$ is normal, so the Urysohn lemma gives a continuous function $g:\overline{V}\to [0,1]$ equal to $1$ on $K$ and $0$ on $\overline{V}\setminus V$. Extend $g$ by zero outside $\overline{V}$. The boundary values make the extension continuous, and its support is a compact subset of $U$.

Fix $x$ and $\epsilon >0$. Since $\mathcal{K}$ is compact in the uniform norm, choose a finite $\epsilon /3$-net $f_1,\dots ,f_r$. By continuity, there is a neighbourhood $U$ of $x$ on which $|f_i(y)-f_i(x)|<\epsilon /3$ for every $i$. For $f$ in $\mathcal{K}$, choose $f_i$ with $\Vert f-f_i{\Vert }_{\infty }<\epsilon /3$. The triangle inequality gives $|f(y)-f(x)|<\epsilon$ on $U$, uniformly in $f$. Thus $\mathcal{K}$ is equicontinuous.

Apply Arzelà-Ascoli theorem on $[-m,m]$. Pointwise boundedness and equicontinuity give a subsequence converging uniformly there. Successively extract subsequences for $m=1,2,\dots$, and take the diagonal subsequence. It converges uniformly on every closed bounded interval. Its limits on nested intervals agree, so they define $f$ on the real numbers; the locally uniform limit of continuous functions is continuous.

A family $F$ of continuous functions on a compact space $K$ is equicontinuous if, for every $x$ and every $\epsilon >0$, there is a neighbourhood $U$ of $x$ such that $|f(y)-f(x)|<\epsilon$ for every $y$ in $U$ and every $f$ in $F$. The Arzelà-Ascoli theorem says that $F$ has compact closure in $C(K)$ with the uniform norm if and only if it is equicontinuous and pointwise bounded.

Attaching the cone collapses the circle $A=S^1\times \{x\}$ in the torus. The quotient has the homotopy typeEquivalently, the exact sequence in part (b) kills one of the two generators of $H_1(T^2)$ and preserves the fundamental two-cycle. Therefore

Write $X^{\prime }=X{\cup }_{A}CA$. After replacing the two pieces by small open neighborhoods, their intersection deformation retracts to $A$. The Mayer-Vietoris sequence, together with $H_i(CA)=0$ for $i>0$ and $H_0(CA)=\mathbb{Z}$, becomesand at degree zero ends aswhich is the required sequence.

Triangulate $X$ and adjoin one new vertex joined to every simplex; the resulting simplicial complex realizes the cone $CX$. The cone contracts to its apex, so

A directed path of length two is determined by its middle vertex $x$ and an ordered pair of distinct neighbors, givingFor the colored complete graph, letA monochromatic triangle contributes three monochromatic two-edge angles to $S$, while a nonmonochromatic triangle contributes one. If $M$ and $N$ are their respective numbers, thenEliminating $N$ gives

The diagonal Ramsey number $R(k)$ is the least $n$ such that every red-blue coloring of $K_n$ contains a monochromatic $K_k$. The standard recursiongivesThus the converted $k^4$ is a superscript-order error; that bound is false for large $k$.

In a red-blue coloring, if the red spanning graph is connected it has a red spanning tree. Otherwise its red components are joined pairwise by blue edges, making the blue graph connected, so it has a blue spanning tree.

The result fails for three colors. Color the six edges of $K_4$ by its three perfect matchings, one color per matching. Every monochromatic graph then consists of two disjoint edges and has no spanning tree.

The map $a\mapsto \{a\}$ injects $x$ into $\mathcal{P}(x)$. Cantor's diagonal argument shows that no map $x\to \mathcal{P}(x)$ is surjective: for $f:x\to \mathcal{P}(x)$, the set $D=\{a:a\notin f(a)\}$ differs from every $f(a)$. Hence $\kappa <2^{\kappa }$.

If $f:x\to y$ is surjective, inverse image gives an injectionFor every infinite initial ordinal ${\omega }_{\alpha }$, transfinite recursion constructs a pairing of ${\omega }_{\alpha }\times {\omega }_{\alpha }$ with ${\omega }_{\alpha }$: at each stage fewer than ${\aleph }_{\alpha }$ earlier pairs have been used. Thuswithout invoking choice for arbitrary families.

Every ordinal below ${\omega }_{\alpha +1}$ has cardinal at most ${\aleph }_{\alpha }$ and can be coded by a well-ordering relation on a subset of ${\omega }_{\alpha }$. Sending such a relation to its order type, and all non-well-orders to zero, gives a surjectionUsing the pairing above to code relations as subsets of ${\omega }_{\alpha }$ yieldsAgain, strictness beyond this is independent of the usual axioms.

For cardinals, $\kappa <\lambda$ means that there is an injection $\kappa \to \lambda$ but no bijection. Under the axiom of choice, every set is well-orderable, so every infinite cardinal is the initial ordinal ${\aleph }_{\alpha }$ for a unique ordinal $\alpha$. Cantor theorem givesand minimality of the successor cardinal givesThe strict inequality ${\aleph }_{\alpha +1}<2^{{\aleph }_{\alpha }}$ printed in the conversion is not provable and can be false under the generalized continuum hypothesis; the intended symbol must be $\le$.