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A topological space is Hausdorff when any two distinct points have disjoint open neighbourhoods.

Any two nonempty open sets of the infinite cofinite topology intersect: their complements are finite sets, so the complement of their intersection is a finite union of finite sets and cannot be all of $Z$. Thus $Z$ is a hyperconnected space.

If the continuous map $f$ took distinct values at $s,t\in Z$, choose disjoint open neighbourhoods $G,H$ of $f(s),f(t)$ in the Hausdorff space $Y$. By continuity, $f^{-1}(G)$ and $f^{-1}(H)$ would be nonempty disjoint open sets of $Z$, a contradiction. ThereforeThe proof establishes the general fact that continuous maps from an infinite cofinite space to a Hausdorff space are constant.

Write $Q=(Z\times I)/\mathcal{R}$. Since $Q$ is homeomorphic to the circle, it is Hausdorff. For fixed $t\in I$, the functionis continuous, being a coordinate slice of the quotient map. Part (a)(i) therefore makes it constant. Choose $x_0\in Z$ and setThen $g$ is continuous andSince the quotient map $p$ is a surjection, so is $g$. This is Hausdorff factorization through a cofinite product factor: a continuous map into a Hausdorff space cannot distinguish the points of the coordinate with the cofinite topology.

If $g$ were injective, it would be a continuous bijection from the compact closed interval $I$ to the Hausdorff space $Q$. Part (a)(ii) would make it a homeomorphism, contradicting the fact that the circle and interval are not homeomorphic.

To recall that obstruction explicitly, remove the interior point $1/2$ from $I$. The resulting topological subspace is separated by the nonempty relatively open sets $[0,1/2)$ and $(1/2,1]$, so it is not connected. The circle with any one point removed is connected: after a rotation, it is the image under a continuous map of the connected open interval $(0,2\pi )$ under $t\mapsto (\cos t,\sin t)$. A homeomorphism and its restriction preserve connectedness, giving a contradiction.

Thus $g$ is not injective. Choose distinct $a,b\in I$ with $g(a)=g(b)$. ThenIn particular the requested $x,a,b$ exist. The argument proves the stronger conclusion that the same two interval coordinates work for every $x$.

Fix $x_0,y_0\in Z$. For a continuous function $f:Z\times Z\to \mathbb{R}$, every coordinate slice is continuous, because the maps $x\mapsto (x,y)$ and $y\mapsto (x_0,y)$ into the product topology are continuous. Their target $\mathbb{R}$ is Hausdorff. Part (a)(i) givesTherefore every such $f$ is a constant function. This is the two-factor case of continuous maps from products of cofinite spaces to Hausdorff spaces are constant.

The second coordinate projection ${\pi }_2:Z\times \mathbb{R}\to \mathbb{R}$ is continuous and nonconstant. If a homeomorphism $H:Z\times Z\to Z\times \mathbb{R}$ existed, ${\pi }_2\circ H$ would be a nonconstant continuous function, contradicting the preceding conclusion. Hence

For the half-plane, put $H_{+}=[0,\infty )$ as a set, so $K=\mathbb{R}\times H_{+}$. Its two ambient topologies give different answers:In $Z\times Z$, if $K$ were closed, its inverse image under the continuous map $y\mapsto (x_0,y)$ would be closed in $Z$. That inverse image is $H_{+}$, which is neither a finite set nor all of $Z$, so it is not closed in the cofinite topology. Thus $K$ is not closed.

To establish compactness, note that the subspace topology on any subset $A$ of a topological space with the cofinite topology is again cofinite. An open cover of a nonempty such $A$ has a nonempty member omitting at most finitely many points; finitely many more members cover those points. Hence every subset of a cofinite space is compact. In particular both $Z$ and $H_{+}$ with its cofinite subspace topology are compact. The subspace topology on $K\subseteq Z\times Z$ is their product topology, as one checks by tracing basis rectangles. The product of two compact spaces is compact, so $K$ is compact here.

For clarity, the finite-product compactness result needs no Hausdorff condition. Given an open cover of $A\times B$, cover each point by a basis rectangle lying in one covering member. For fixed $a\in A$, use compactness of $B$ to select finitely many rectangles whose second factors cover $B$, and intersect their first factors to get an open neighbourhood $U_a$ of $a$. Those finitely many original members cover $U_a\times B$. Finally finitely many $U_a$ cover $A$, producing a finite subcover of $A\times B$.

In $Z\times \mathbb{R}$, the second coordinate projection hasSince the ray is closed in the usual real line, $K$ is closed. It is not compact: its image under a continuous map under ${\pi }_2$ is $[0,\infty )$, which is not compact. For an explicit obstruction, the open intervals $(-1,n)$, $n\ge 1$, cover this ray but have no finite subcover.

Finally consider the diagonal $W$ in $Z\times \mathbb{R}$. DefineThis is a bijection. The identity function from the usual real line to $Z$ is continuous: each nonempty cofinite open set is the complement of a finite set, and finite sets are closed in the usual real line. Both coordinates of $\phi$ are therefore continuous, so $\phi$ is continuous for the product topology and the subspace topology of $W$. Its inverse is the restricted second coordinate projection, which is continuous. ThereforeFor the second conclusion, $Z$ is compact whereas $W\cong \mathbb{R}$ is not compact; homeomorphisms preserve compactness. The diagonal in a product with cofinite and Euclidean topologies thus inherits the usual topology, despite one cofinite factor.

Any two nonempty open sets of the infinite cofinite topology intersect: their complements are finite sets, so the complement of their intersection is finite and cannot be all of $\mathbb{R}$.

Suppose the quotient space had distinct points with disjoint open neighbourhoods $U,V$. Since the quotient map $p$ is surjective, both preimages $p^{-1}(U),p^{-1}(V)$ would be nonempty. Since $p$ is continuous, these would be disjoint open subsets of $\mathbb{R}$ with its cofinite topology, contradicting the observation above. Thus any such quotient space with at least two points is not Hausdorff.

A quotient space of a space carrying the cofinite topology has a topology contained in the cofinite topology on its underlying quotient set. If $q:X\to Y$ is the quotient map and $U$ is a nonempty open subset of $Y$, its preimage has finite complement. Since $q$ is surjective, the complement of $U$ is the image of that finite complement, hence is finite. Equality need not hold: partition $\mathbb{Z}$ into its even and odd integers. Both classes are infinite, so their two-point quotient space has the indiscrete topology, whereas the cofinite topology on a finite set is the discrete topology.

By convergence of a sequence in a topological space, $x_n\to a$ means that every open neighbourhood of $a$ contains all sufficiently late terms. In the cofinite topology, this is equivalent to the condition that, for every $b\ne a$, the value $b$ occurs only finitely often. Necessity uses the open neighbourhood $X\setminus \{b\}$; sufficiency chooses an index beyond every last occurrence of the finitely many points excluded by a given open neighbourhood.

Consequently a sequence with distinct terms converges to every point of $X$. For the original choice $x_n=n$ on $X={\mathbb{N}}_{>0}$, take the two open neighbourhoodsAll terms with $n\ge 9$ lie in $U_1$ and all terms with $n\ge 8$ lie in $U_2$. More generally, each finite excluded set has a last visit, so $x_n\to 1$ and $x_n\to 2$, and in fact $x_n\to a$ for every $a\in X$. No numerical distance is tending to zero here; the open sets specify the meaning of convergence.

Distinct terms are sufficient, not necessary. The sequence $1,1,2,2,3,3,\dots$ also converges to every point, since each value occurs only twice. By contrast, the constant sequence $x_n=1$ converges only to $1$: an open neighbourhood of any $a\ne 1$ can exclude $1$. The alternating sequence $1,2,1,2,\dots$ converges to no point, since some value different from any proposed limit occurs infinitely often.

A topological space is sober when every nonempty irreducible closed subset is the closure of a unique point. Every Hausdorff space is sober. In contrast, an infinite set with the cofinite topology has the whole space irreducible and closed, while every singleton is closed, so it is not sober.

A topological space is T1 when every point is a closed point. Equivalently, for distinct $x,y$ there is an open set containing $x$ but not $y$, and an open set containing $y$ but not $x$. These open sets need not be disjoint. Every Hausdorff space is T1, but an infinite set with its cofinite topology is T1 without being Hausdorff.

These sets form a topology: a nonempty union of upper rays is another upper ray, or all of $\mathbb{R}$ when their endpoints have no lower bound; a finite intersection of rays has endpoint the maximum of their endpoints. The whole line must be included. The continuous map $n\mapsto n$ from $\mathbb{N}$ with the cofinite topology to this space is nonconstant, because each upper ray has cofinite inverse image.