Munkres, Topology, 2nd ed. — constructing topological spaces, then the concepts built upon them
Chapter 1 established the set-theoretic vocabulary; Chapter 2 begins the subject proper. It defines what a topological space is (§12), and then devotes most of its length to methods of constructing topologies: from a basis (§13), from an order (§14), as a product of two spaces (§15), and as a subspace of an existing space (§16). It then turns to the concepts a topology exists to support — closed sets and limit points (§17), and continuous functions (§18), the central object of the theory. The remaining sections generalize the product construction (§19), recover the metric spaces of real analysis as a special case (§20–21), and introduce quotient spaces (§22).
Sections §13 through §16 each answer a version of the same question: given some raw material — a basis, an order, two existing spaces, a subset — how is a legitimate topology obtained from it? Every answer must be checked against the three axioms of §12, which is why that section precedes the rest and must be known precisely.
The sections below are built out in full, with worked examples, warnings, quiz questions, and practice problems. Later sections will be added in the same form as the material is reached.
The definition everything else in the book depends on
A topology on a set $X$ is a collection $\mathcal T$ of subsets of $X$, called the open sets, satisfying three conditions. The definition took decades to settle into its present form: it needed to be broad enough to include every space of real mathematical interest — Euclidean space, infinite-dimensional space, function spaces — while narrow enough that the familiar theorems about those spaces would continue to hold in general. What survived is three short conditions, and the entire subject is built on top of them.
(1) $\varnothing$ and $X$ are both in $\mathcal T$.
(2) The union of the elements of any subcollection of $\mathcal T$ is in $\mathcal T$ — arbitrarily many open sets, even uncountably many, union together to something open.
(3) The intersection of the elements of any finite subcollection of $\mathcal T$ is in $\mathcal T$.
$(X, \mathcal T)$ is then called a topological space (often just written $X$, with $\mathcal T$ left implicit). A set $U$ counts as "open" exactly when $U \in \mathcal T$ — openness isn't a property a set has on its own, it's a status conferred by whichever topology you've chosen.
Unions may be taken over any subcollection of $\mathcal T$, however large; intersections are guaranteed only for finite subcollections. This asymmetry is essential to the definition, not an oversight. A concrete instance of why it cannot be relaxed already appears in Modern Analysis: $G_n = (-1/n, 1/n) \subset \mathbb R$ is open for every $n$, yet $\bigcap_{n=1}^\infty G_n = \{0\}$, which is not open. Were axiom (3) to permit infinite intersections, the real line with its ordinary topology would fail to be a topological space at all. This is the standard example to produce whenever a problem asks for a case where an infinite intersection of open sets fails to be open.
Each interval $(-1/n,\,1/n)$ sits strictly inside the previous one. Their union is simply the widest row shown, $n=1$, since every other row is already contained in it. Their intersection shrinks as $n$ grows and, taken over all $n$, collapses to the single point $\{0\}$ — a point has no open ball around it, so the intersection is not open.
Establishing that a given collection is a topology means checking all three axioms explicitly, in order. The two examples below carry out a full verification, and then examine a collection that fails, so that the point of failure is visible directly.
(1) $\varnothing \in \mathcal T$ ✓ and $X \in \mathcal T$ ✓ — both listed explicitly.
(2) Check every union: $\{1\}\cup\{1,2\}=\{1,2\}\in\mathcal T$ ✓. Any union involving $X$ or $\varnothing$ trivially stays in $\mathcal T$. All unions land back in the collection. ✓
(3) Check intersections: $\{1\}\cap\{1,2\}=\{1\}\in\mathcal T$ ✓. All other intersections trivially work. ✓
Conclusion: $\mathcal T$ is a topology on $X$. It is, in relabeled form, the example in the upper-right corner of Munkres Figure 12.1.
(1) $\varnothing, X \in \mathcal T$ ✓.
(2) Check unions: $\{1\}\cup\{2\} = \{1,2\}$. Is $\{1,2\} \in \mathcal T$? No — it's not on the list.
Axiom (2) fails: two sets of $\mathcal T$ have a union that does not itself belong to $\mathcal T$. This collection is therefore not a topology. Adding $\{1,2\}$ to $\mathcal T$ — precisely the topology verified above — repairs it. Declaring a set of singletons open obligates the inclusion of every union they generate as well.
These four constructions recur throughout the book as standard test cases. Given a claimed property of topological spaces, these are the first spaces to check it against, since together they span the extremes of how many sets can be open.
$\mathcal T = \mathcal P(X)$: every subset is open. This is the finest possible topology on $X$. Every singleton $\{x\}$ is open, which forces a great deal of additional structure — in the discrete topology every function out of $X$ is automatically continuous, and every subset is both open and closed.
$\mathcal T = \{\varnothing, X\}$: only the two sets every topology is required to contain. This is the coarsest possible topology. Almost nothing is open, and consequently almost nothing can be distinguished — no two distinct points can be separated by open sets.
$U \in \mathcal T_f$ iff $X - U$ is finite, or $U = \varnothing$. This is the first topology in the book whose validity is not immediate, and the verification of axiom (3) is instructive: if $U_1,\dots,U_n \in \mathcal T_f$ are nonempty, then $X - \bigcap U_i = \bigcup (X-U_i)$, a union of finitely many finite sets, hence finite. The restriction to finitely many terms is exactly what axiom (3) requires, and exactly what makes the argument work.
The same construction one level up: $U \in \mathcal T_c$ iff $X-U$ is countable, or $U=\varnothing$. Axiom (2) requires a countable union of countable sets to remain countable, which holds. The general principle — arbitrary unions of open sets are always open, while only finite intersections are guaranteed to be — is what does the work in every one of these examples; countability plays a supporting role here, not the decisive one.
Once you accept that a single set $X$ can carry many different topologies, the natural next question is how to compare them.
If $\mathcal T'\supset \mathcal T$, we say $\mathcal T'$ is finer than $\mathcal T$ (strictly finer if the containment is proper), and $\mathcal T$ is coarser than $\mathcal T'$. If either containment holds, the two topologies are comparable.
Munkres compares a topology to a truckload of gravel, with the open sets as pebbles. Breaking the pebbles into smaller pieces produces more of them, and a topology built from smaller pieces is correspondingly finer, in the sense that finer gravel means smaller stones. The discrete topology is the limiting case of dust; the indiscrete topology, a single boulder.
Two topologies on the same set need not be comparable: neither containment need hold. Comparability is a partial order among the topologies on $X$, and most pairs of topologies on a given set are simply incomparable.
A small generating collection that produces an entire topology automatically
Listing an entire topology directly is usually unworkable — the standard topology on $\mathbb R$ already has uncountably many open sets. A basis lets a much smaller collection of "generating" sets stand in for the whole topology, with the rest produced automatically.
A collection $\mathcal B$ of subsets of $X$ (the basis elements) is a basis for a topology on $X$ if:
(1) For each $x \in X$, at least one basis element contains $x$.
(2) If $x$ belongs to the intersection of two basis elements $B_1, B_2$, there is a basis element $B_3$ with $x \in B_3 \subset B_1 \cap B_2$.
Given such a $\mathcal B$, declare $U \subset X$ open exactly when: for every $x \in U$, some basis element $B$ satisfies $x \in B \subset U$. This collection of open sets is the topology generated by $\mathcal B$.
Condition (2) does not require $B_1 \cap B_2$ itself to be a basis element — only that some basis element fits inside the overlap, snugly around $x$. This is a weaker, more flexible requirement, and it is exactly the property that lets unions of basis elements survive intersection, which is the whole content of the next section.
Checking the three axioms against this generation rule is short, and condition (2) is used at exactly one step.
Axiom (1): $\varnothing$ is open vacuously — there are no points to check. $X$ is open because condition (1) of the basis definition already puts every point into some basis element, and every basis element is trivially $\subset X$.
Axiom (2), arbitrary unions: if $x \in \bigcup_\alpha U_\alpha$, then $x \in U_\alpha$ for some particular $\alpha$; since $U_\alpha$ is open, some basis element fits inside it around $x$, and that same basis element sits inside the whole union. A single witness suffices, exactly as in the general case.
Axiom (3), finite intersections — where condition (2) is used directly: suppose $x \in U_1 \cap U_2$. Openness of $U_1$ gives a basis element $B_1$ with $x \in B_1 \subset U_1$; openness of $U_2$ gives $B_2$ with $x \in B_2 \subset U_2$. Now $x \in B_1 \cap B_2$, and condition (2) of the basis definition hands over a third basis element $B_3$ with $x \in B_3 \subset B_1 \cap B_2 \subset U_1 \cap U_2$ — exactly the certificate needed to conclude the intersection is open. Without condition (2), this step has no way to proceed.
The topology generated by $\mathcal B$ equals the collection of all unions of elements of $\mathcal B$.
So a basis is a set of building blocks, and the topology is everything assemblable from them by gluing arbitrarily many together — nothing more is needed. Unlike a basis in linear algebra, this representation is not unique: the same open set can typically be written as a union of basis elements in more than one way.
These range from fully geometric to purely combinatorial, and the second pair makes a point worth sitting with: a basis is never unique — different-looking generating collections can produce the identical topology.
Let $\mathcal B$ be every open circular region in $\mathbb R^2$. Condition (1) is immediate. Condition (2): given two overlapping disks and a point $x$ in their overlap, a smaller disk fits around $x$ entirely inside the lens-shaped overlap — geometrically clear, since the overlap of two disks is itself an open region with room around $x$. This generates the ordinary topology on $\mathbb R^2$: $U$ is open iff every point of $U$ has some disk around it inside $U$.
Let $\mathcal B'$ be every open rectangle with sides parallel to the axes. Condition (2) is now trivial rather than merely true: the intersection of two axis-aligned rectangles is itself a rectangle (or empty), so no third, smaller element is ever needed. Remarkably, $\mathcal B'$ generates the same topology as the disks — any disk can be covered from inside by rectangles, and any rectangle by disks. A basis is a choice of generating set, not a canonical, uniquely-determined object.
$\mathcal B = \{(a,b) : a < b\}$, the collection of all open intervals, generates what is called the standard topology on $\mathbb R$ — the topology meant whenever $\mathbb R$ is discussed without further qualification, including everywhere earlier in this chapter. Condition (2) holds for the same reason as the rectangles: the intersection of two open intervals is again an open interval, or empty.
Let $\mathcal B$ be every one-point subset $\{x\}$ of $X$. Condition (1) is trivial. Condition (2): two distinct singletons never intersect, so the condition holds vacuously (there is no point $x$ in $B_1 \cap B_2 = \varnothing$ to produce a $B_3$ for). The topology generated is every possible union of singletons — every subset of $X$ whatsoever — exactly the discrete topology.
Section 13.01 goes from a basis to a topology. The reverse direction is needed constantly: given a topology already in hand, how do you tell whether some smaller collection happens to be a basis for it? This is exactly the tool used later, in §16, to show that a basis for $X$ restricts to a basis for a subspace $Y$.
Let $X$ be a topological space. Suppose $\mathcal C$ is a collection of open sets of $X$ such that for every open set $U$ and every $x \in U$, some $C \in \mathcal C$ satisfies $x \in C \subset U$. Then $\mathcal C$ is a basis for the topology of $X$.
This is precisely the criterion invoked when checking that $\mathcal B_Y = \{B \cap Y : B \in \mathcal B\}$ is a basis for the subspace topology in §16: given $U$ open in $X$ and $y \in U \cap Y$, an element $B$ of the original basis with $y \in B \subset U$ gives $y \in B\cap Y \subset U \cap Y$ directly — exactly the hypothesis of this lemma.
Since a topology is often specified through a basis rather than listed outright, it helps to have a way of testing fineness directly from the bases, without reconstructing either full topology.
Let $\mathcal B$ and $\mathcal B'$ be bases for topologies $\mathcal T$ and $\mathcal T'$ on $X$. Then $\mathcal T'$ is finer than $\mathcal T$ if and only if: for every $x \in X$ and every basis element $B \in \mathcal B$ containing $x$, there is a basis element $B' \in \mathcal B'$ with $x \in B' \subset B$.
The gravel analogy resolves this directly: think of $\mathcal B$'s elements as pebbles, and $\mathcal B'$'s elements as the dust those pebbles get smashed into. The finer topology's dust particles fit inside the coarser topology's pebbles — smaller elements, more of them, finer topology. That inclusion, $B' \subset B$, is exactly the direction Lemma 13.3 states.
Two further topologies on $\mathbb R$, both strictly finer than the standard one, generated by different bases:
Lower limit topology $\mathbb R_\ell$: generated by all half-open intervals $[a,b) = \{x : a \le x < b\}$.
$K$-topology $\mathbb R_K$: generated by all open intervals $(a,b)$ together with all sets $(a,b) - K$, where $K = \{1/n : n \in \mathbb Z^+\}$.
Both are strictly finer than the standard topology: given a basis element $(a,b)$ of the standard topology and a point $x$ in it, $[x,b)$ is a basis element of $\mathbb R_\ell$ with $x \in [x,b) \subset (a,b)$ — the inclusion Lemma 13.3 asks for. But no ordinary open interval around $x$ can shrink to fit inside $[x,d)$, since every open interval containing $x$ spills over to the left of $x$, which $[x,d)$ does not — so the reverse inclusion fails, and the fineness is strict.
$\mathbb R_\ell$ and $\mathbb R_K$, however, are not comparable with one another: each is finer than the standard topology in its own way, but neither basis's elements can be uniformly shrunk to fit inside the other's. This is a concrete instance of the fact from §12.04 that fineness is only a partial order — most pairs of topologies are simply incomparable, and this is the first non-trivial pair encountered so far.
How any subset of a topological space becomes a topological space in its own right
Given a topological space $(X,\mathcal T)$ and any subset $Y \subset X$ — not assumed open, closed, or anything else — there's a canonical way to turn $Y$ itself into a topological space, using only the topology $X$ already has.
$\mathcal T_Y$ is a topology on $Y$, called the subspace topology; $Y$ with this topology is a subspace of $X$. Its open sets are exactly the intersections of $X$'s open sets with $Y$.
Axiom (1): $\varnothing = Y \cap \varnothing$ and $Y = Y \cap X$, and $\varnothing, X \in \mathcal T$, so both land in $\mathcal T_Y$.
Axioms (2)/(3): intersection with $Y$ distributes over unions and intersections: $(U_1\cap Y)\cap\cdots\cap(U_n\cap Y) = (U_1\cap\cdots\cap U_n)\cap Y$, and $\bigcup_\alpha (U_\alpha \cap Y) = \big(\bigcup_\alpha U_\alpha\big)\cap Y$. Since $\mathcal T$ is already closed under finite intersections and arbitrary unions, intersecting the $U_\alpha$ with $Y$ first does not disturb that closure. The verification reduces entirely to these two set identities.
Once $Y$ carries its own topology, the word "open" is no longer unambiguous: a set can be open in $Y$ without being open in $X$. This distinction accounts for most errors made when first working with subspaces, and every claim about openness in this setting should specify relative to which space it is made.
Let $X=\mathbb R$ (usual topology), $Y=[0,1]$. Is $[0,\,0.5)$ open in $Y$? Yes: $[0,0.5) = (-1,\,0.5)\cap Y$, and $(-1,0.5)$ is open in $\mathbb R$. But $[0,0.5)$ is not open in $\mathbb R$ itself — no open interval around the point $0$ fits inside it. The same set is open in one space and not in the other, depending on which topology is in question.
If $U$ is open in $Y$, and $Y$ itself is open in $X$, then $U$ is also open in $X$.
Proof: $U$ open in $Y$ means $U = Y\cap V$ for some $V$ open in $X$. If $Y$ is also open in $X$, then $Y\cap V$ is an intersection of two sets open in $X$, hence open in $X$ by axiom (3).
The hypothesis that $Y$ itself is open in $X$ is essential: in the example above, $[0,1]$ is not open in $\mathbb R$, which is exactly why the conclusion of the lemma does not apply, and $[0,0.5)$ remains open in $Y$ without being open in $\mathbb R$.
This lemma depends on the notion of a basis, treated fully in §13: a basis $\mathcal B$ for a topology is a smaller generating collection of sets, with the property that every open set is a union of basis elements — so that specifying $\mathcal B$ specifies the whole topology.
If $\mathcal B$ is a basis for the topology of $X$, then $\mathcal B_Y = \{B\cap Y \mid B \in \mathcal B\}$ is a basis for the subspace topology on $Y$.
The argument mirrors the definition of the subspace topology itself: intersecting each generating set with $Y$ again produces a generating collection, now for the subspace topology on $Y$.
A natural question follows: when $Y$ arises from some other construction on $X$ — a product, an order — does the subspace topology on a further subset agree with the topology that construction produces directly? For products, it does.
If $A$ is a subspace of $X$ and $B$ is a subspace of $Y$, then the product topology on $A\times B$ is the same as the topology $A\times B$ inherits as a subspace of $X\times Y$.
The general basis element for $X\times Y$ is $U\times V$ ($U$ open in $X$, $V$ open in $Y$), and $(U\times V)\cap(A\times B) = (U\cap A)\times(V\cap B)$, which is exactly the general basis element for the product topology on $A\times B$. The two constructions therefore generate the same basis, and hence the same topology.
Unlike the product case, restricting an order topology to a subset does not, in general, reproduce the subspace topology on that subset. The order topology on $Y$ and the subspace topology on $Y$ are two independently defined topologies, and there is no guarantee in advance that they coincide.
In the subspace topology, $\{2\}$ is open: it equals $(\tfrac32,\tfrac52)\cap Y$. But in the order topology on $Y$ (treating $Y$ as an ordered set with the inherited order), $\{2\}$ is not open — every order-topology basis element containing $2$ has the form $\{x\in Y : a < x \le 2\}$ for some $a\in Y$, and every such set also contains points of $Y$ less than $2$, so it can never shrink down to just $\{2\}$. The two topologies on $Y$ are each well defined, and they disagree about this particular set.
Call $Y\subset X$ convex in $X$ if for every $a
Intervals and rays are always convex, which is why $[0,1]$, having no gaps, presented no difficulty, while $[0,1)\cup\{2\}$, with a gap between $1$ and $2$, is exactly where the two topologies diverge. The gap is what permits the divergence: the missing interval $[1,2)$ is where the order topology would otherwise supply additional open sets, and their absence allows a point such as $2$ to be isolated in the subspace topology in a way the order topology has no means of reproducing.
Before treating the subspace and order topologies on a subset $Y$ of an ordered space as identical, check whether $Y$ is convex — an interval or ray, with no gap between two of its own points. If it is, the two topologies coincide by Theorem 16.4. If it is not, they must be checked directly, and any divergence will appear at the boundary of the gap.
§12, §13, and §16 — select an answer for immediate, explained feedback
§12, §13, and §16 — attempt each problem before consulting the hints or the solution