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Reading Munkres' text on Topology, we get the fairly straight-forward definition of a basis:

$\mathcal{B}$ is a basis for a topology on $X$ if $\mathcal{B}$ is a collection of subsets of $X$ 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 elements $B_1$ and $B_2$, then there us a basis element $B_3$ containing $x$ such that $B_3 \subset B_1 \cap B_2$.

Now, Munkres proceeds to (roughly) define the topology $\tau$ generated by $\mathcal{B}$ contains elements $U$ so that for each $x \in U$ there is a basis element $B \in \mathcal{B}$ such that $x \in B$ and $B \subset U$.

Consider the set $X = \{a,b,c\}$. Then, by definition, $\mathcal{B} = \{\{a\},\{b\},\{c\}\}$ is a basis for a topology on $X$. We proceed to (attempt to) find the topology generated by $\mathcal{B}$. Clearly, $\{a\},\{b\},\{c\} \in \tau$. We note that given our definitions, the topology $\tau$ generated by $\mathcal{B}$ is $\{X, \emptyset, \{a\},\{b\},\{c\},\{a,b\},\{a,c\},\{b,c\}\}$.

Is this correct, or have I misunderstood something? In particular, does this mean that we may have bases of different cardinalities?

tryst with freedom
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Andrew Thompson
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  • That's a bit confused. $\tau$ must contain subsets of $X$, $\mathcal P(X)$ is not a subset of $X$. – Thomas Andrews Aug 24 '14 at 17:09
  • Did you mean the topolog $\tau$ generated by $\mathcal{B}$ is $\mathcal{P}(X)$? – Quang Hoang Aug 24 '14 at 17:09
  • Oh, blunder with $\mathcal{P}(X)$ – Andrew Thompson Aug 24 '14 at 17:11
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    @AndrewThompson The new answer is correct, but you could just as well have said: $\tau=\mathcal P(X)$. Any topology in which every singleton is an open set has every set an open set. That topology is called the "discrete topology." – Thomas Andrews Aug 24 '14 at 17:16
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    A related interesting example: The family of all open intervals $(a, b)$ forms a basis for the usual topology on $\Bbb R$. The family of open intervals with rational endpoints $(p, q)$ where $p,q\in \Bbb Q$ also forms a basis for the usual topology in $\Bbb R$. But the second basis is countable while the first is uncountable. – MJD Aug 25 '14 at 15:50
  • That is indeed very interesting. I will play around with that. – Andrew Thompson Aug 25 '14 at 17:07

1 Answers1

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You misunderstood something. Since $\tau$ only includes subsets of $X$. It does not include $\mathcal P(X)$ itself as an element.

Instead it includes every element of $\mathcal P(X)$ (read: subset of $X$) which can be written as the union of these singletons. Which subsets are these?  

Yes, this is correct. Every subset of $X$ is open in this case.

As for the final question, yes, it is possible to have bases of different cardinalities, for example by taking $\{\{a\},\{b\},\{c\},\{a,b\}\}$ you obtain another basis for the same topology, and of course the topology itself is always a basis for itself.

Asaf Karagila
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  • For the "As for the final question,..." part, one can note that the topology itself is a basis. – Quang Hoang Aug 24 '14 at 17:12
  • I edited my question, as I blundered with the notation. Is it correct now? (In my hand-written notes I had "stuff = $\mathcal{P}(X)$") – Andrew Thompson Aug 24 '14 at 17:14
  • Interesting. However, there are coarser topologies than this. For example, consider the following topology on $X$: $\tau = {X, \emptyset, {a}}$. It is not possible to create a basis for $X$ that generates this topology..well, given your confirmation, I should be able to figure out the rest by myself. Thank you! – Andrew Thompson Aug 24 '14 at 17:29
  • @Andrew: It is possible. Remember that $X$ and $\varnothing$ are always added to the topology (where $\varnothing$ can be seen trivially as a union of no sets; $X$ is sometimes required to be in the basis, or the union of all the elements of the basis, but we can also require it to always be added explicitly, since it has to be there anyway. Munkers seems to follow the second convention regarding to $X$). – Asaf Karagila Aug 24 '14 at 17:36
  • Ah, of course, we do not necessarily need the singletons in our basis. Thanks again! – Andrew Thompson Aug 24 '14 at 17:45
  • I don't understand, why did you cross it out instead of deleting the text? – tryst with freedom Feb 26 '22 at 19:52