One of the problem I’m facing while doing topology from James Munkres book is that everything(definition, theorem, proof) is written in words, instead of symbols and quantifiers. Sometimes which(words) makes things ambiguous.

The following is the definition of basis:

If $X$ is a set, a basis for a topology on $X$ is a collection $\mathscr{B}$ of subsets of $X$ (called basis elements) such that

(1)For each $x∈X$, there is at least one basis element $B$ containing $x$

(2)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 $B_3⊆B_1∩B_2$.

If $\mathscr{B}$ satisfies these two conditions, then we define the topology $\mathfrak{I}$ generated by $\mathscr{B}$ as follows : A subset $U$ of $X$ is said to be open in $X$( i.e., to be an element of $\mathfrak{I}$ ) if for each $x\in U,$ there is a basis element $B\in \mathscr{B}$ such that $x\in B$ and $B\subset U.$

Above definition could have been written in much elegant and concise way.

Question: (1) “for a topology on X” - Why are we using the word topology? When we don’t know anything about the topology or topological space? We can just defined a set, we’ll call basis, if that set satisfy (1) & (2). Later based on this set(which have some structure), we’ll define topology. Summary: Given $\mathscr{B} \subseteq \wp(X)$(power set), check if $\mathscr{B}$ satisfy two given conditions and if $\mathscr{B}$ satisfy those conditions, then we’ll call set $\mathscr{B}$ a basis, without mentioning any topology on $X$.

(2) In this book, Mukres uses the word “open set” lots of times. What if we have more than one topology on $X$, then $U$ is open, means what?

The trend follows (3)

Lemma 13.1. Let $X$ be a set; let $\mathscr{B}$ be a basis for a topology $\mathfrak{I}$ on $X$. Then $\mathfrak{I}$ equals the collection of all unions of elements of $\mathscr{B}$.

Now what does “$\mathscr{B}$ be a basis for a topology $\mathfrak{I}$ on $X$” means? Is it topology $\mathfrak{I}$ generated from basis $\mathscr{B}$? If yes, then why not use that wording? Certainly you would agree that we can’t prove the lemma with this “ collection of all unions of elements of $\mathscr{B}$” wording. One need to change from word format to symbols format to prove the lemma. How do i do it?

  • $\begingroup$ It seems to me a simple "variation" of the usual definition of Topology: "A topological space is an ordered pair $(X,\tau)$ where $X$ is a set and $\tau$ is a collection of subsets of $X$ such that... the collection $\tau$ is called a topology on $X$." $\endgroup$ Dec 22, 2021 at 13:56
  • $\begingroup$ If I hear a word like "basis" then immediately a question arises in my mind: "a basis of what?..." I think Munkres is aware that phenomenom and immediately provides an answer already ("..of a topology.."). No problem for me. $\endgroup$
    – drhab
    Dec 22, 2021 at 14:01
  • $\begingroup$ @MauroALLEGRANZA yeah. The members(all) of $\mathfrak{I}$ are called open. If we have $\gt 1 $ topology on $X$, then I think a set $U$ is open don’t make sense. Because it don’t tell $U$ is in which topology. $\endgroup$
    – user264745
    Dec 22, 2021 at 14:01
  • $\begingroup$ One rarely considers more than one topology on a set at a time. If you do, then you just say something like "The set $U$ is open with respect to $\mathfrak{I}_1$ but not with respect to $\mathfrak{I}_2$." $\endgroup$
    – Dan Rust
    Dec 22, 2021 at 14:03
  • $\begingroup$ @DanRust ohhh... that’s nice. $\endgroup$
    – user264745
    Dec 22, 2021 at 14:05

2 Answers 2


A topology on $X$ (a set), as you know, is just a set of subsets of $X$ satisfying certain axioms.

Sometimes we can just decsribe this collection in one fell swoop (like the cofinite one, the discrete and the trivial one), but often we define a topology via a basis: a (hopefully smaller and more managable collection of subsets $\mathcal{B}$ of $X$) such that a set is open (i.e. in the topology) iff we can write it as some union of elements from $\mathcal{B}$. This leads to often more workable definitions of open sets, i.e. topologies.

The conditions he presents are necessary and sufficient for a collection $\mathcal{B}$ so that the description of a topology as unions from it actually makes sense and is valid.

So he intends to define a topology using $\mathcal{B}$, but then you first have to check these two conditions on it (for which you only have to know $\mathcal{B}$ itself, and $X$ and no more). After that being done yoy can say you define a topology via the unions (this is what he calls "generating the topology" from $\mathcal{B}$).

OTOH he considers the opposite situation: you already some way are in possession of a topology $\mathcal{T}$ on $X$ and wonder could I have generated it by some base $\mathcal{B}$ as well? Well, $\mathcal{B} = \mathcal{T}$ works trivially, but we'd like a smaller collection somehow that generates $\mathcal{T}$. So he considers $\mathcal{B}$ a base for the topology $\mathcal{T}$ iff

$\mathcal{B} \subseteq \mathcal{T}$ and for every $O \in \mathcal{T}$ and every $x \in O$ we have some $B_x \in \mathcal{B}$ so that $x \in B_x \subseteq O$.

And the lemma you stated in fact says that (in this case) if we had chosen $\mathcal{B}$ as a base on $X$ then $\mathcal{T}$ would have been exactly the generated topology by $\mathcal{B}$. So the two ways of being a base mean the same in the end.

  • $\begingroup$ Thank you for the answer. I like your 4th paragraph from top. Btw what is OTOH? I somewhat understand your OTOH paragraph, but why not just be more concrete. Like, show If $\mathfrak{I} _{\mathscr{B}}$ (denotes topology generated by basis $\mathscr{B}$), then it’s equal to $\{ \cup_{\alpha \in A} B_{\alpha} | B_{\alpha} \in \mathscr{B}$ & A is indexing set$\}$. We already know one definition of $\mathfrak{I} _{\mathscr{B}} =\{ U \subseteq X|\forall x\in U,\exists B\in \mathscr{B}$ s.t $x\in B\subseteq U \}$. Now show there is a equivalent definition(Union) of $\mathfrak{I} _{\mathscr{B}}$. $\endgroup$
    – user264745
    Dec 22, 2021 at 17:48
  • $\begingroup$ @user264745 OTOH = on the other hand. $\endgroup$ Dec 22, 2021 at 17:50
  • $\begingroup$ Ok. Is whatever I said in the comment correct? $\endgroup$
    – user264745
    Dec 22, 2021 at 17:52
  • 1
    $\begingroup$ @user264745 yes, in essence. If the candidate subfamily of the topology must generate the topology it’s clear itmust satisfy the $B_x$ condition. But then it also satisfies 1 and 2, and starting from the base we get the topology back, as stated. $\endgroup$ Dec 22, 2021 at 17:56

Proof of Lemma 13.1.

If $U\in\mathfrak I$ then for every $x\in U$ there is some $B_x\in\mathscr B$ with $x\in B_x\subseteq U$ so that we can write:$$U=\bigcup_{x\in U}B_x$$showing that $U$ can be written as a union of elements of $\mathscr B$.

Conversely every element of $\mathscr B$ is also an element of topology $\mathfrak I$. A topology is closed under unions so that a union of elements of $\mathscr B$ will again be an element of $\mathfrak I$.

  • $\begingroup$ Thank you for the answer. I think proof of this lemma is quite standard. We need to show to sets are equal, i.e. $\mathfrak{I} _{\mathscr{B}}$(denotes the topology generated by basis $\mathscr{B}$)= $\{ \cup_{\alpha \in A} B_{\alpha} | B_{\alpha} \in \mathscr{B}$ & A is indexing set$\}$. Proof of this lemma, is just simply to show both sets are subset of each other, basic set theory. $\endgroup$
    – user264745
    Dec 22, 2021 at 17:18

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