Confusion Regarding Munkres's Definition of Basis for a Topology The definition of Basis for a Topology as given in Munkres's book is as follows,

If $X$ is a set, a basis for a topology on $X$ is a collection $\mathcal{B}$ of subsets of $X$ (called basis elements) such that
  
  
*
  
*For each $x∈X$, there is at least one basis element $B$ containing $x$
  
*If $x$ belongs to the intersection of two basis elements $B_1$ and $B_2$, then there is a basis
  element $B_3$ containing $x$ such that $B_3⊆B_1∩B_2$.

My question is,

In the definition though we are defining "basis for a topology", there is no mention of the topology on $X$. What role the topology on $X$ plays in the definition? Is it just a simple printing mistake or I am missing something?

 A: A bit further in the book he will discuss how such a basis on a set generates a basis for a topology on that set by taking unions of basis elements.

Conversely, if you start with a space $X$ which is already topologized, then a collection $\mathcal{B}$ of open subsets such that any open subset of $X$ can be written as a union of elements of $\mathcal{B}$ is a basis for the set $X$ in the sense of the definition above. (And, via this basis, we recover the topology on $X$.)
Proof. We check the first condition. Let $x \in X$. Since $X$ is open as a subset of itself, it may be written as a union of basis elements, $X = \bigcup_\mathcal{B} B$. Therefore by the definition of union, $x$ belongs to some $B \in \mathcal{B}$.
Now for the second condition. If $x$ belongs to the two open sets $B_1,B_2 \in \mathcal{B}$, it must also belong to the open set $B_1 \cap B_2$. Writing this set as a union of basis elements, $B_1 \cap B_2 = \bigcup_\mathcal{B} B$, we see again that by the definition of union there must exist some basis element $B_3 \subset B_1 \cap B_2$ containing $x$.
A: You don't need a topology to define a base for a topology, but every basis generates a topology, and you can go about finding bases for existing topologies.
The topology generated by such a basis is determined in the following way:
A subset $\mathcal O$ of $X$ is open if for all $x\in \mathcal O$, there is an element $\mathcal B$ of the basis such that $x\in \mathcal B\subseteq \mathcal O$.
Expressed another way, you can say that the open subsets are the ones which are unions of basis elements.
Why use bases? Well, just like in liner algebra, it's a way to compress information about all open sets into a smaller collection that recovers everything else. For example, the open balls of rational radius around points in $\mathbb R^2$ form a countable basis of the ordinary topology on $\Bbb R^2$. This collection is not closed under intersection or union, and the full topology on $\mathbb R$ contains uncountably many open sets.
A: This is a completely revised version of my previous answer, which was not correct. The existence of a major error in my previous answer was recently pointed out to me by Merk Zockerborg.
The main result relevant to the question asked by user 170039 is Corollary 2 below.
Contents


*

*Basis for a topological space $(X,\tau).$

*Given ${\mathcal B} \subseteq {\cal P}(X),$ find a topology ${\tau}_{\mathcal B}$ on $X$ such that $\mathcal B$ is a basis for $(X,\tau_{\mathcal B}).$

*Comparing $\tau$ and $\tau_{\mathcal B}$ when $(X,\tau)$ is a topological space and ${\mathcal B} \subseteq {\cal P}(X)$ is such that $(X,\tau_{\mathcal B})$ is a topological space.
Notation: $X$ is a set and ${\cal P}(X)$ is the collection of all subsets of $X.$ The (sometimes subscripted) symbols $\tau$ and $\mathcal B$ will denote subsets of ${\cal P}(X).$ Most of the time (but not always) $\tau$ will be a topology on $X$ and $\mathcal B$ will be a basis for a topology on $X.$
In the spirit of making this an expository overview of some introductory ideas that arise when introducing the notion of a basis for a topological space, I am not including proofs of the stated theorems. However, those actively studying this topic in topology may find it useful to provide proofs.
1. Basis for a topological space $(X,\tau).$

Definition: Let $(X,\tau)$ be a topological space. A basis for $(X,\tau)$ is a collection $\mathcal B$ of subsets of $X$ such that:
(a) $\;\;\mathcal B \subseteq \tau$
(b) $\;$ For each $x \in X,$ and for each $U \in \tau$ such that $x \in U,$ there exists $V \in \mathcal B$ such that $x \in V$ and $V \subseteq U.$

Note: The term "base" is more commonly used than "basis", but I'll use "basis" in agreement with the usage by user 170039.
Roughly speaking, if we consider (open) neighborhoods of $x$ as "measures of being close to $x$", where being closer to $x$ is described by the use of a smaller (in the subset sense) neighborhood of $x,$ then a basis for $\tau$ is a subcollection of open sets that is sufficient to describe being arbitrarily close to any given point.
Note that this notion of being close to $x$ has more structure to it than a notion entirely based on the subset relation, because we also have the property that any finite number of "close to $x$ conditions" can be replaced by a single "close to $x$ condition", due to the finite intersection property of open sets. More specifically, being simultaneously $U_1$-close to $x$ and $U_2$-close to $x$ can be replaced by being $(U_1 \cap U_2)$-close to $x$ in the case of $\tau,$ and by being $V$-close to $x$ for some $\mathcal B$-element $V$ such that $V \subseteq U \overset{\text{def}}{=} U_1 \cap U_2$ in the case of $\mathcal B.$ For example, suppose $X = \{x,y,z\}$ and $\tau = \{\;\{x,y\}, \; \{y,z\}\;\}.$ Note that $\tau$ is not a topology on $X.$ Nonetheless, we can still talk about being close to elements of $X.$ For example, there is the notion of being "$\{x,y\}$-close to $y$" and there is the notion of being $\{y,z\}$-close to $y.$ However, note that there is no single closeness condition that implies each of these conditions, because the only subset of $X$ containing $y$ that is a subset of $\{x,y\}$ and a subset of $\{y,z\}$ is $\{y\},$ and $\{y\} \not \in \tau.$
A topology can have more than one basis. For example, in the case of the real numbers with its usual topology, the collection of all open intervals of finite length is a basis and the collection of all open intervals of finite length with rational endpoints is a basis.
Question: [This occurred to me when I was writing these notes, and I thought others here might be interested.] How many bases exist for the usual topology of the real numbers? The answer is $2^{c}.$ Since each such basis is a collection of open sets, and there are $2^c$ many collections of open sets of $\mathbb R$ (because there are $c$ many open sets of ${\mathbb R}),$ it follows that there are at most $2^c$ many bases for the usual topology of the real numbers. The following shows that there are at least $2^c$ many bases for the usual topology of the real numbers. Let $D_1$ and $D_2$ be disjoint subsets of ${\mathbb R}$ such that $D_1$ has cardinality $c$ and $D_2$ is dense-in-${\mathbb R}$ (for example, $D_1$ could be the set of irrational numbers between $0$ and $1,$ and $D_2$ could be the set of rational numbers), and consider the collection of open intervals each of whose endpoint(s) belongs to $D_1 \cup D_{2}.$ This collection of open intervals is a basis for ${\mathbb R}.$ Moreover, if we remove from this collection any set consisting only of open intervals each of whose endpoint(s) belongs to $D_{1},$ then what remains will also be a basis, and there are $2^c$ many ways to remove such sets of intervals from the original collection of open intervals (because there are $2^c$ many subsets of the collection of open intervals each of whose endpoint(s) belongs to $D_{1}).$

Theorem 1: Let $(X,\tau)$ be a topological space and let $\mathcal B$ be a collection of subsets of $X$ such that:
(a) $\;\;\mathcal B \subseteq \tau$
(b) $\;$ Every element of $\tau$ is the union of some subcollection of elements in $\mathcal B.$
Then $\mathcal B$ is a basis for $(X,\tau).$

Regarding (b) above, we are using the convention that an empty union of sets is possible, and that an empty union of sets is equal to the empty set.

Theorem 2: Let $(X,\tau)$ be a topological space and let $\mathcal B$ be a basis for $(X,\tau).$ Then every element of $\tau$ is the union of some subcollection of elements in $\mathcal B.$

Theorems 1 and 2 together imply that we could have defined "$\mathcal B$ is a basis for $(X,\tau)$" by replacing (b) in our earlier definition with (b) in Theorem 1. This alternate way of defining a basis for $(X,\tau)$ is used in some books.
Note that (b) in Theorem 1 provides a way to generate all elements of $\tau$ from the elements in $\mathcal B$ — take unions. This is analogous to having a way to generate all elements in a vector space from a (vector space) basis — take linear combinations.
2. Given ${\mathcal B} \subseteq {\cal P}(X),$ find a topology ${\tau}_{\mathcal B}$ on $X$ such that $\mathcal B$ is a basis for $(X,\tau_{\mathcal B}).$
Suppose that no topology on the set $X$ has been provided. Can we obtain a topology ${\tau}_{\mathcal B}$ on $X$ by picking some collection $\mathcal B \subseteq {\cal P}(X)$ and letting ${\tau}_{\mathcal B}$ be the collection of all possible unions of elements in $\mathcal B$?

Example 1: Let $X = \{x,y\}$ and let $\mathcal B = \{\; \{x\}\;\}.$ Then the collection of all possible unions of elements in $\mathcal B$ gives ${\tau}_{\mathcal B} = \{\; \emptyset,\;\{x\}\;\},$ but ${\tau}_{\mathcal B}$ is not a topology on $X,$ because $X \notin {\tau}_{\mathcal B}.$ (A simpler example is to use $\mathcal B = \emptyset,$ but I wanted to give a less trivial example.)

Perhaps we can fix the problem that arises in Example 1 by assuming $\cup {\mathcal B} = X.$ Note that $\cup {\mathcal B} = X$ is equivalent to the first bullet assumption in user 170039’s statement of the definition of "basis for a topology" from Munkres's book. Example 2 shows that having $\cup {\mathcal B} = X$ is not sufficient for ${\tau}_{\mathcal B}$ to be a topology on $X.$

Example 2: Let $X = \{x,y,z\}$ and let ${\mathcal B} = \{\;\{x,y\},\;\{y,z\}\;\}.$ Then $\cup {\mathcal B} = X.$ However, the collection of all possible unions of elements in $\mathcal B$ gives ${\tau}_{\mathcal B} = \{\;\emptyset,\; \{x,y\},\;\{y,z\},\; X\;\},$ and ${\tau}_{\mathcal B}$ is not a topology because ${\tau}_{\mathcal B}$ is not closed under finite intersections — note that $\{x,y\} \cap \{y,z\} = \{y\}$ and $\{y\} \notin {\tau}_{\mathcal B}.$

It turns out that fixing both problems, namely the problem that Example 1 illustrates and the problem that Example 2 illustrates, is sufficient for ${\tau}_{\mathcal B}$ to be a topology on $X.$

Theorem 3: Let $X$ be a set and let $\mathcal B$ be a collection of subsets of $X$ such that:
(a) $\;\; \cup {\mathcal B} = X$
(b) $\;$ For each positive integer $n,$ if $B_1 \in \mathcal B$ and $B_2 \in \mathcal B$ and $\cdots$ and $B_n \in {\mathcal B},$ then $B_1 \cap B_2 \cap \cdots \cap B_n$ can be written as the union of (possibly infinitely many) elements in ${\mathcal B}.$
If we let ${\tau}_{\mathcal B}$ be the collection of all possible unions of elements in $\mathcal B,$ then $(X,{\tau}_{\mathcal B})$ is a topological space and $\mathcal B$ is a basis for $(X,{\tau}_{\mathcal B}).$

In some books Theorem 3 is buried in results about a subbasis for a topological space. Explicit statements of it can be found in Theorem 3.50 on p. 26 of Introduction to General Topology by Helen Frances Cullen (1968) and in 2.2 Base Characterization on p. 153 of An Introduction to Topology and Homotopy by Allan John Sieradski (1992). On the off-chance that anyone looks at Cullen's book, note that in her book she assumes an empty intersection gives the underlying universal set (see top of p. 416). Since this is not a standard assumption, I've avoided this assumption by including (a).
Theorem 4 is the more commonly stated condition for a collection of subsets of $X$ to generate a topology on $X,$ where the method of generation is by using the collection of all possible unions of elements in that collection of subsets. At the risk of stating the obvious, if $\mathcal B$ satisfies the assumptions in Theorem 3 and $\mathcal B$ also satisfies the assumptions in Theorem 4 (in fact, one can show that satisfying the assumptions in either theorem implies satisfying the assumptions in the other theorem), then the topology obtained using Theorem 3 is the same as the topology obtained using Theorem 4. This follows from the fact that in each theorem the same procedure is used to obtain the topology, namely the topology is the collection of all possible unions of elements in ${\mathcal B}.$

Theorem 4: Let $X$ be a set and let $\mathcal B$ be a collection of subsets of $X$ such that:
(a) $\;\; \cup {\mathcal B} = X$
(b) $\;$ If $U \in \mathcal B$ and $V \in \mathcal B$ and $x \in U \cap V,$ then there exists $W \in \mathcal B$ such that $x \in W$ and $W \subseteq U \cap V.$
If we let ${\tau}_{\mathcal B}$ be the collection of all possible unions of elements in $\mathcal B,$ then $(X,{\tau}_{\mathcal B})$ is a topological space and $\mathcal B$ is a basis for $(X,{\tau}_{\mathcal B}).$

One the things that (b) in Theorem 4 does is to incorporate our earlier observation (long paragraph a little below the definition of "basis") that finitely many "close to $x$" conditions can be replaced by a single "close to $x$" condition. Note that in Example 2, ${\tau}_{\mathcal B}$ has the property that "$\{x,y\}$-close to $y$ and $\{y,z\}$-close to $y$" cannot be equaled or strengthened by any single ${\tau}_{\mathcal B}$-closeness notion.
In our earlier vector space analogy, Theorems 3 and 4 are somewhat akin to finding conditions under which a set of vectors is a linearly independent set, then defining the linear span of those vectors, and finally observing that the linear span of those vectors forms a vector space.
3. Comparing $\tau$ and $\tau_{\mathcal B}$ when $(X,\tau)$ is a topological space and ${\mathcal B} \subseteq {\cal P}(X)$ is such that $(X,\tau_{\mathcal B})$ is a topological space.
Let $(X,\tau)$ be a topological space and let $\mathcal B$ be a collection of subsets of $X$ satisfying the assumptions in either Theorem 3 or Theorem 4 (or both). In this situation we have two topological spaces to contend with, $(X,\tau)$ and $(X,\,\tau_{\mathcal B}).$ In general, we have $\tau \neq \tau_{\mathcal B}.$ Indeed, there is no reason why we would expect there to be any relation between $\tau$ and $\tau_{\mathcal B},$ aside from the fact that each is a collection of subsets of $X$ that satisfies the axioms for a topology on $X.$

Example 3: Suppose ${\mathcal B} \not \subseteq {\tau}.$ Then it is easy to see that $\tau \neq \tau_{\mathcal B}.$ (Proof: Let $B \in \mathcal B$ such that $B \not \in {\tau}.$ Then $B$ belongs to $\tau_{\mathcal B},$ because we always have ${\mathcal B} \subseteq \tau_{\mathcal B},$ and hence $B$ does not belong to ${\tau}.$ Therefore, $\tau \neq \tau_{\mathcal B}.)$

Example 4, due to Merk Zockerborg, shows that ${\mathcal B} \subseteq {\tau}$ is not sufficient to conclude that $\tau = \tau_{\mathcal B}.$

Example 4: Let $X = \{x,y\}$ and $\tau = \{\;\emptyset,\;\{x\},\;\{y\},\;\{x,y\}\;\}$ and ${\mathcal B} = \{\;\{x\},\;\{x,y\}\;\}.$ Then ${\mathcal B} \subseteq {\tau}$ and $\tau_{\mathcal B} = \{\;\emptyset,\;\{x\},\;\{x,y\}\;\}.$ Therefore, $\tau \neq \tau_{\mathcal B}.$

Corollary 2 below gives one possible assumption that, along with the assumption ${\mathcal B} \subseteq {\tau},$ is sufficient to conclude that $\tau = \tau_{\mathcal B}.$

Theorem 5: Let $X$ be a set and let ${\mathcal B}_1,$ ${\mathcal B}_2$ each be collections of subsets of $X$ that satisfy the assumptions in either Theorem 3 or Theorem 4. Then ${\tau}_{{\mathcal B}_1} \subseteq {\tau}_{{\mathcal B}_2}$ if and only if the following assertion holds:
For each $B_1 \in {\mathcal B}_1$ and for each $x \in B_{1},$ there exists $B_2 \in {\mathcal B}_2$ such that $x \in B_2$ and $B_2 \subseteq B_{1}.$

$\;$

Corollary 1: Let $X$ be a set and let ${\mathcal B}_1,$ ${\mathcal B}_2$ each be collections of subsets of $X$ that satisfy the assumptions in either Theorem 3 or Theorem 4. Then ${\tau}_{{\mathcal B}_1} = {\tau}_{{\mathcal B}_2}$ if and only if both of the following assertions hold:
For each $B_1 \in {\mathcal B}_1$ and for each $x \in B_{1},$ there exists $B_2 \in {\mathcal B}_2$ such that $x \in B_2$ and $B_2 \subseteq B_{1}.$
  For each $B_2 \in {\mathcal B}_2$ and for each $x \in B_{2},$ there exists $B_1 \in {\mathcal B}_1$ such that $x \in B_1$ and $B_1 \subseteq B_{2}.$

$\;$

Corollary 2: Let $(X,\tau)$ be a topological space and let $\mathcal B$ be a collection of subsets of $X$ that satisfies the assumptions in either Theorem 3 or Theorem 4 such that:
(a) $\;\; {\mathcal B} \subseteq {\tau}$
(b) $\;$ For each $U \in \tau$ and for each $x \in U,$ there exists $B \in \mathcal B$ such that $x \in B$ and $B \subseteq U.$
Then $\tau = \tau_{\mathcal B}.$

Proof of Corollary 2: From ${\mathcal B} \subseteq {\tau}$ and the fact that $\tau$ is closed under arbitrary unions, it follows that $\tau_{\mathcal B} \subseteq {\tau}.$ Applying Theorem 5 with ${\mathcal B}_1 = \tau$ and ${\mathcal B}_2 = {\mathcal B},$ and observing that ${\tau}_{{\mathcal B}_1} = {\tau}_{\tau} = \tau$ (because $\tau$ is closed under arbitrary unions), we get $\tau \subseteq \tau_{\mathcal B}.$ (The notation is slightly confusing, since in ${\tau}_{\tau}$ the subscripted $\tau$ is a topology on $X$ and the other $\tau$ is part of the notation that symbolizes the result of carrying out a certain operation on subscripted set.)
