Topological Spaces and Continuity: Examples and Counterexamples

Let $X$ be a set, let $\mathcal{T}_{\mathrm{disc}} = \mathcal{P}(X)$ be the discrete topology and $\mathcal{T}_{\mathrm{ind}} = \{\varnothing, X\}$ the indiscrete topology (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies), and let $A \subseteq X$. Then:

1. In the discrete space every subset is clopen, and $\operatorname{int}(A) = A = \overline{A}, \qquad \partial A = \varnothing$ for every $A$ (Interior, closure, boundary, exterior, derived set and isolated point in a topological space). The singletons $\{\, \{x\} : x \in X \,\}$ form a basis (Basis and subbasis for a topology, and the topology generated by a family of sets).
2. In the indiscrete space $\operatorname{int}(A) = \begin{cases} X & A = X \\ \varnothing & A \ne X \end{cases}, \qquad \overline{A} = \begin{cases} \varnothing & A = \varnothing \\ X & A \ne \varnothing \end{cases},$ so $\partial A = X$ for every $A$ other than $\varnothing$ and $X$.
3. Maps out of a discrete space and into an indiscrete space are all continuous. For any topological space $Y$, every function $(X, \mathcal{T}_{\mathrm{disc}}) \to Y$ is continuous, and every function $Y \to (X, \mathcal{T}_{\mathrm{ind}})$ is continuous.
4. The other two directions are restrictive. A function $f : Y \to (X, \mathcal{T}_{\mathrm{disc}})$ is continuous exactly when $f^{-1}[\{x\}]$ is open in $Y$ for every $x \in X$; and a function $g : (X, \mathcal{T}_{\mathrm{ind}}) \to Y$ is continuous exactly when $g^{-1}[V] \in \{\varnothing, X\}$ for every open $V \subseteq Y$.

The two topologies are the extreme points of the comparison order (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison): every topology on $X$ is finer than $\mathcal{T}_{\mathrm{ind}}$ and coarser than $\mathcal{T}_{\mathrm{disc}}$.

• Both spaces are first countable, for opposite reasons. In the discrete space $\{\{x\}\}$ is a one-element neighbourhood base at $x$; in the indiscrete space $\{X\}$ is. So first countability is no obstruction to either, and it is the Hausdorff property that separates them: the discrete topology is metrizable by the metric taking the value $1$ on distinct points and $0$ on equal ones, whose ball of radius $1$ about $x$ is $\{x\}$, while the indiscrete topology on a set with two or more points is metrizable by none (The indiscrete topology on a two-point set is induced by no metric).

• Claim 3 explains why neither extreme is interesting on its own. A space in which every map is continuous carries no information about the maps; the content of a topology lies between the two extremes, and the comparison order of Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison is the scale on which that content is measured.

• In the indiscrete space every nonempty subset is dense by claim 2 and Dense, nowhere dense and codense subsets of a topological space, and the criterion by basic open sets, and in the discrete space only $X$ is. These are again the two extremes: density is a measure of how coarse the topology is, not of how large the set is.

Let $X$ be an infinite set with the cofinite topology $\mathcal{T}_{\mathrm{cof}}$, whose open sets are $\varnothing$ together with the sets of finite complement (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies, Finite, countably infinite, countable, uncountable). Then:

1. Closures. For $A \subseteq X$, $\overline{A} = \begin{cases} A & A \text{ finite} \\ X & A \text{ infinite,} \end{cases} \qquad \operatorname{int}(A) = \begin{cases} A & X \setminus A \text{ finite} \\ \varnothing & X \setminus A \text{ infinite.} \end{cases}$
2. A subset is dense if and only if it is infinite (Dense, nowhere dense and codense subsets of a topological space, and the criterion by basic open sets); in particular every infinite subset is dense, and no finite subset is.
3. No two nonempty open sets are disjoint. If $U, V \in \mathcal{T}_{\mathrm{cof}}$ are nonempty then $U \cap V \ne \varnothing$.
4. Every singleton is closed, so points are distinguishable by closed sets; nevertheless claim 3 says distinct points are never separated by disjoint open sets, so the space is as far from Hausdorff as a space with closed points can be.

• Claim 3 is the reason the cofinite topology is a standard counterexample factory. A space with at least two points in which any two nonempty open sets meet cannot be metrizable (Distinct points of a metric space have disjoint balls around them, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not), since a metric separates distinct points by disjoint balls. It does not follow that sequential limits fail to be unique there: the cocountable topology on $\mathbb{R}$ also has no two disjoint nonempty open sets and its sequential limits are unique (In the cocountable topology on $\mathbb{R}$ the closed sets are the countable sets and $\mathbb{R}$, and a sequence converges iff it is eventually constant). In the cofinite topology uniqueness does fail, but the argument needs an injective sequence and not merely the meeting of open sets (In the indiscrete topology every sequence converges to every point, and in the cofinite topology on an infinite set an injective sequence converges to every point).

• Finiteness of $X$ collapses the example. If $X$ were finite then every subset would have finite complement, the topology would be discrete, and all four claims would read differently or vacuously. The hypothesis that $X$ is infinite is used in steps 1.5 and 2.2 and is not decoration.

• The same computation with "at most countable" in place of "finite" gives the cocountable topology, whose behaviour on $\mathbb{R}$ is the subject of the next example (In the cocountable topology on $\mathbb{R}$ the closed sets are the countable sets and $\mathbb{R}$, and a sequence converges iff it is eventually constant). The two differ in exactly one respect that matters here: on $\mathbb{R}$ the cocountable topology still has closed points and no two disjoint nonempty open sets, but its convergent sequences are far more restricted.

Give $\mathbb{R}$ the cocountable topology $\mathcal{T}_{\mathrm{coc}}$, whose open sets are $\varnothing$ together with the sets whose complement is at most countable (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies, Finite, countably infinite, countable, uncountable). Then:

1. The closed sets are exactly the at most countable subsets of $\mathbb{R}$ together with $\mathbb{R}$ itself, and these two families are disjoint, $\mathbb{R}$ being uncountable ($\mathbb{R}$ is uncountable (Cantor's nested intervals, 1874)). In particular every singleton is closed.
2. Closures. For $A \subseteq \mathbb{R}$, $\overline{A} = \begin{cases} A & A \text{ at most countable} \\ \mathbb{R} & A \text{ uncountable.} \end{cases}$
3. A sequence converges if and only if it is eventually constant (Convergence and cluster points of a sequence in a topological space, sequential continuity, and the sequential closure), and then it converges to its eventual value and to no other point.

Claim 3 is what makes this space the standard witness that sequences can be blind to a topology: the convergent sequences are the same as in the discrete topology, while the topology itself is very far from discrete by claim 2.

• Assuming the Axiom of Countable Choice, this space is not first countable. Under that hypothesis Assuming Countable Choice, in a first countable space sequential closure equals closure and sequential continuity at a point equals continuity there would otherwise force $\operatorname{seqcl}(A)=\overline{A}$ for every $A$, whereas claim 3 makes the sequential closure of $[0,1]$ equal to $[0,1]$ and claim 2 makes its closure all of $\mathbb{R}$ (In the cocountable topology on $\mathbb{R}$ the sequential closure of $[0,1]$ is $[0,1]$ while its closure is all of $\mathbb{R}$). Under the same hypothesis it is therefore not metrizable either.

• No two nonempty open sets are disjoint here either. If $U$ and $V$ are nonempty and open then $\mathbb{R} \setminus (U \cap V)$ is a union of two at most countable sets and hence at most countable, so $U \cap V$ cannot be empty, $\mathbb{R}$ being uncountable ($\mathbb{R}$ is uncountable (Cantor's nested intervals, 1874)). The argument is the one used for the cofinite topology (On an infinite set the cofinite topology has every infinite subset dense and no two nonempty open sets disjoint) with "at most countable" in place of "finite".

• The topology is strictly coarser than the discrete topology and strictly finer than the cofinite topology on $\mathbb{R}$. It is strictly coarser than discrete because $\{0\}$ is not open, its complement being uncountable ($\mathbb{R}$ is uncountable (Cantor's nested intervals, 1874), Every subset of an at most countable set is at most countable). It is finer than cofinite because a finite set is at most countable, and strictly so because $\mathbb{R} \setminus \mathbb{N}$ has an at most countable complement that is not finite.

Let $X$ be a set, let $p \in X$, and give $X$ the particular-point topology $\mathcal{T}_p = \{\varnothing\} \cup \{\, U \subseteq X : p \in U \,\}$ (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies). Then, for $A \subseteq X$:

1. Closed sets. The closed subsets of $(X, \mathcal{T}_p)$ are $X$ together with the sets not containing $p$.
2. Interior and closure (Interior, closure, boundary, exterior, derived set and isolated point in a topological space): $\operatorname{int}(A) = \begin{cases} A & p \in A \\ \varnothing & p \notin A, \end{cases} \qquad \overline{A} = \begin{cases} X & p \in A \\ A & p \notin A. \end{cases}$ In particular $\{p\}$ is dense (Dense, nowhere dense and codense subsets of a topological space, and the criterion by basic open sets) and is the smallest dense subset, while $X \setminus \{p\}$ has empty interior.
3. No two nonempty open sets are disjoint, since every nonempty open set contains $p$.
4. Sierpinski space is the case of a two-point set. With $S = \{a,b\}$, $a \ne b$, and particular point $b$, the topology is $\mathcal{T}_{\mathrm{Sier}} = \{\varnothing, \{b\}, S\}$; here $\overline{\{b\}} = S$ and $\overline{\{a\}} = \{a\}$, so $b$ is the open point and $a$ the closed point.
5. Continuous maps into Sierpinski space are exactly the open subsets of the source. For a topological space $Y$, the assignment $f \mapsto f^{-1}[\{b\}]$ is a bijection from the set of continuous maps $Y \to (S, \mathcal{T}_{\mathrm{Sier}})$ onto the topology of $Y$ (For a map of spaces the following agree: continuity at every point, preimages of open sets open, preimages of closed sets closed, preimages of subbasic open sets open, and $f(\overline{A}) \subseteq \overline{f(A)}$).

• Sierpinski space is the smallest space that is not indiscrete and not discrete, and claim 5 is why it matters: it represents the notion "open set" as a mapping problem, in the same way that a two-element set represents "subset". Every topology on $Y$ is recovered as the set of continuous maps $Y \to S$.

• When $X$ has at least two points, the particular-point topology separates distinct points only in the weakest sense. Any two distinct points are distinguished by an open set: if one of them is $p$, then $\{p\}$ is open and contains $p$ but not the other; and if neither is $p$, then $\{x,p\}$ is open and contains $x$ but not $y$. It is not Hausdorff: every two nonempty open sets meet at $p$, and $\{p\}$ is dense by claim 2. When $X=\{p\}$, by contrast, the topology is discrete and the separation axioms hold vacuously. In every case the space is first countable, since $\{\{x,p\}\}$ is a one-element neighbourhood base at $x$: any neighbourhood of $x$ contains an open set containing $x$, which contains $p$ as well.

• A comparison with the cofinite topology. When $X$ is infinite, both have the property that any two nonempty open sets meet (On an infinite set the cofinite topology has every infinite subset dense and no two nonempty open sets disjoint), but the particular-point topology achieves it with a single point doing all the work. When $X$ has at least two points, its particular point $p$ is not closed, whereas every point is closed in the cofinite topology.

Let $(L, \le)$ be a totally ordered set (Partial order and partially ordered set) with at least two elements. For $a \in L$ write

$L_{<a} := \{\, t \in L : t < a \,\}, \qquad L_{>a} := \{\, t \in L : a < t \,\}$

for the open rays, and let $\mathcal{S}_L$ be the family of all open rays. The order topology on $L$ is $\mathcal{T}_{\mathrm{ord}} := \langle \mathcal{S}_L \rangle$, the topology generated by $\mathcal{S}_L$ (Basis and subbasis for a topology, and the topology generated by a family of sets). Then:

1. A basis. By A family is a basis for a unique topology iff it covers the set and every point of an intersection of two members lies in a member inside that intersection; finite intersections of any subbasis form a basis the finite intersections of open rays form a basis for $\mathcal{T}_{\mathrm{ord}}$, and every such intersection is $L$ itself (the empty intersection), an open ray, or an open interval $(a,b) := L_{>a} \cap L_{<b}$. So $\mathcal{B}_L := \{L\} \cup \mathcal{S}_L \cup \{\, (a,b) : a, b \in L \,\}$ is a basis for the order topology.
2. On $\mathbb{R}$ the order topology is the usual topology. Taking $L = \mathbb{R}$ with its order (Order on the reals, Ordered field), $\mathcal{T}_{\mathrm{ord}} = \mathcal{T}_{d_{\mathbb{R}}}$, the metric topology of $d_{\mathbb{R}}(x,y) = |x-y|$ (The absolute value makes $\mathbb{R}$ a metric space: $d(x,y) = |x-y|$ is a metric, its open balls are the intervals $(x-r, x+r)$, and it is unbounded, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not). The rays and intervals of claim 1 are then exactly the intervals of that shape in the sense of Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length.

Claim 2 identifies the order topology of $\mathbb{R}$ with a topology already in the library rather than introducing a second one.

• All three descriptions of $\mathbb{R}$'s topology name one collection of open sets. Claim 2 identifies the order topology with the metric topology of $d_{\mathbb{R}}$, and Which results on this page use the order of $\mathbb{R}$ and therefore have no general-topological analogue records that the metric topology and the order-native topology built earlier in this library are in turn the same collection. Nothing below uses that third description; it is named so that a reader moving between the pages knows there is one topology and not two.

• The hypothesis that $L$ has at least two elements is what keeps the rays from being useless: on a one-point set every ray is empty and the order topology is the only topology there is. Nothing else in claim 1 uses it.

• The order topology is not always metrizable, and the order alone does not decide the matter. Claim 2 is a statement about $\mathbb{R}$ and is proved from the specific fact that the balls of $d_{\mathbb{R}}$ are the bounded open intervals; no general theorem is being invoked, and none is available here.

• The Sorgenfrey line is not the order topology of the usual order on $\mathbb{R}$ (The Sorgenfrey line: $\mathbb{R}$ with the half-open intervals $[a,b)$ as a basis is strictly finer than the usual topology, is first countable, has a countable dense subset, and its sequences converge only from the right). It is generated by the half-open intervals $[a,b)$, which are not unions of open rays and open intervals, so it is strictly finer than the order topology of that order; the order it comes from is the same order, which shows that "generated by intervals" is not the same as "the order topology". Whether some other total order on $\mathbb{R}$ has the Sorgenfrey topology as its order topology is a different question, and nothing here or elsewhere in this library answers it.

Let $\mathcal{B} := \{\, [a,b) : a, b \in \mathbb{R},\ a < b \,\}$ be the family of bounded half-open intervals of $\mathbb{R}$ (Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length). Then:

1. $\mathcal{B}$ is a basis (A family is a basis for a unique topology iff it covers the set and every point of an intersection of two members lies in a member inside that intersection; finite intersections of any subbasis form a basis) for a topology $\mathcal{T}_{\mathrm{S}}$ on $\mathbb{R}$. The space $(\mathbb{R}, \mathcal{T}_{\mathrm{S}})$ is the Sorgenfrey line, also called the lower limit topology.
2. $\mathcal{T}_{\mathrm{S}}$ is strictly finer than the usual topology (Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not, The absolute value makes $\mathbb{R}$ a metric space: $d(x,y) = |x-y|$ is a metric, its open balls are the intervals $(x-r, x+r)$, and it is unbounded): every set open in the usual topology is in $\mathcal{T}_{\mathrm{S}}$, and $[0,1)$ is in $\mathcal{T}_{\mathrm{S}}$ and is not open in the usual topology.
3. The Sorgenfrey line is first countable (First countable space: a countable neighbourhood base at every point): for $x \in \mathbb{R}$ the family $\{\, [x,\ x + 1/(k+1)) : k \in \mathbb{N} \,\}$ is an at most countable neighbourhood base at $x$.
4. It has an at most countable dense subset, namely the rationals: $\mathbb{Q}$ is dense in $(\mathbb{R}, \mathcal{T}_{\mathrm{S}})$ (Dense, nowhere dense and codense subsets of a topological space, and the criterion by basic open sets) and is at most countable ($\mathbb{Q}$ is countably infinite, Finite, countably infinite, countable, uncountable).
5. Sequences converge only from the right. For a sequence $(x_k)$ in $\mathbb{R}$ and $x \in \mathbb{R}$, $x_k \to x$ in $\mathcal{T}_{\mathrm{S}}$ (Convergence and cluster points of a sequence in a topological space, sequential continuity, and the sequential closure) if and only if for every real $\varepsilon > 0$ there is $K \in \mathbb{N}$ with $x \le x_k < x + \varepsilon$ for all $k \ge K$. In particular the sequence $y_k := x - 1/(k+1)$ converges to $x$ in the usual topology and does not converge to $x$ in $\mathcal{T}_{\mathrm{S}}$.

At this point in the reading order, separability has not yet been defined; the later definition Separability: the existence of an at most countable dense subset ↗ abbreviates claim 4.