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I'm learning Topology by Munkres' book and I am a bit confused about the definition he gives (in p.78) of a basis of a topology:

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

  1. For each $x \in X$, there is at least one basis element $B$ containing $x$.

  2. If $x$ belongs to the intersection of two basis element $B_1$ and $B_2$, then there is a basis element $B_3$ containing $x$ such that $B_3 \subset B_1 \cap B_2$.

My question is: shouldn’t Munkres mean $B_3 \subseteq B_1 \cap B_2$? I mean, subset instead of proper subset?

For instance, in Example 2 (Figure 4) below, isn’t the rectangle $B_3$ encountered, the very intersection of the rectangles $B_1$ and $B_2$? enter image description here

Any help would be highly appreciated.

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    Some authors use $\subset$ to mean $\subseteq$ -- check Munkres' definiition of $\subset$ !! Certainly that's what he means here: Otherwise, given a set with the indiscrete topology, the topology itself would not be a basis. – BrianO Oct 18 '15 at 10:05
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    Unfortunately, Munkres uses $\subset$ to mean subset and $\subsetneq$ to mean proper subset. – Henry Swanson Oct 18 '15 at 10:05
  • $\LaTeX$ agrees with Munkres, $\subset$ is \subset, not \propersubset. – Daniel Fischer Oct 18 '15 at 10:20
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    @Daniel: However, the distinction is between \subset and \subseteq, which can just as well be interpreted as distinguishing proper subset from subset, so $\LaTeX$ really isn’t dispositive, even if anyone thought that it actually had much bearing on the subject. Using $\subset$ for $\subseteq$ is just inviting trouble. – Brian M. Scott Oct 18 '15 at 23:29

1 Answers1

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(1)Most writers use both $\subset$ and $\subseteq$ and $\subseteqq$ interchangeably,and use only $\subsetneq$ or $\subsetneqq$ to denote proper subset.Most topologists say "base" ,not "basis".Otherwise it can be confusing when discussing topological vector spaces.In Munkres' def'n, $B_3$ cannot be required to be a proper subset of $B_1\cap B_2$.Example : $X=\{1,2,3\}$ with exactly 5 open sets : $ B_o=\phi , B_1=\{1,2\}, B_2=\{2,3\}, B_3=\{2\},B_4=X.$......(2)Most writers prefer to define a base $B$ for a topology $T$ on $X$ to be any $B\subset T$ (including the case $B=T$) such that any $t\in T$ is a union of a collection of members of $B$. (This includes the case $t=\phi$ because $\phi$ is the union of an empty collection.)They then prove that a family $B$ of subsets of a set $X$ is a base for a topology $T$ on $X$ if and only if $B$ satisfies Munkres' conditions 1. and 2.

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