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I am confused with the concept of topology base. Which are the properties a base has to have?

Having the next two examples for $X=\{a,b,c\}$:

1) $(X,\mathcal{T})$ is a topological space where $\mathcal{T}=\{\emptyset,X,\{a\},\{b\},\{a,b\}\}$. Which is the general procedure to follow in order to get a base for $(X,\mathcal{T})$? Can $\mathcal{B}=\{\{a\},\{b\}\}$ be a base?

2) Having just $X$ and no topology $\mathcal{T}$ defined for $X$, is $\mathcal{A}=\{X,\{a\},\{c\}\}$ a base? What is the topology that it generates?

Thank you very much.

Stefan Hansen
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Haritz
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3 Answers3

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Let $X$ be a non-empty set. A collection $\mathscr{B}$ of subsets of $X$ is a base for some topology on $X$ if it satisfies two conditions:

  1. $\mathscr{B}$ covers $X$. That is, every point of $X$ belongs to at least one member of $\mathscr{B}$.
  2. If $B_1,B_2\in\mathscr{B}$ and $x\in B_1\cap B_2$, then there is a $B_3\in\mathscr{B}$ such that $x\in B_3\subseteq B_1\cap B_2$.

These two conditions are exactly what is needed to ensure that

$$\mathscr{T}=\Big\{\bigcup\mathscr{A}:\mathscr{A}\subseteq\mathscr{B}\Big\}$$

is a topology on $X$. In words, the collection of all unions of members of $\mathscr{B}$ is a topology on $X$, the topology generated by the base.

Note that a topology may have many different bases. The topology $\big\{\varnothing,\{a\},\{b\},\{a,b\}\big\}$ on the set $\{a,b\}$ has the following bases:

  1. $\big\{\varnothing,\{a\},\{b\},\{a,b\}\big\}$
  2. $\big\{\{a\},\{b\},\{a,b\}\big\}$
  3. $\big\{\varnothing,\{a\},\{b\}\big\}$
  4. $\big\{\{a\},\{b\}\big\}$

Conditions (1) and (2) above are the easiest way to characterize the families of sets that are bases for some topology on $X$. If you already have a topology $\mathscr{T}$ on $X$, you can say simply that a subset $\mathscr{B}$ of $\mathscr{T}$ is a base for $\mathscr{T}$ if and only if every member of $\mathscr{T}$ (i.e., every open set in the space $\langle X,\mathscr{T}\rangle$) is a union of members of $\mathscr{B}$.

Yes, $\big\{\varnothing,\{a\},\{c\},X\big\}$ is a base for a topology on $X$: it satisfies the two conditions given at the beginning of this answer. The topology that it generates is

$$\big\{\varnothing,\{a\},\{c\},\{a,c\},\{a,b,c\}\big\}\;.$$

Brian M. Scott
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  • The third base that you propose, {{a},{b}}, should not be {0,{a},{b}}? Because {a}Intersection{b} = 0. And 0 is not in {{a},{b}}.Yes? – Haritz Nov 20 '12 at 10:50
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    @Haritz: No: it is never necessary to include $\varnothing$ in a base. Condition (2) says that if there is a point $x\in B_1\cap B_2$, then there must be a $B_3\in\mathscr{B}$ such that $x\in B_3\subseteq B_1\cap B_2$; it says nothing about the case in which $B_1\cap B_2=\varnothing$. – Brian M. Scott Nov 20 '12 at 10:56
  • Thank you very much @Brian M. Scott – Haritz Nov 20 '12 at 11:01
  • @Haritz: You’re very welcome. – Brian M. Scott Nov 20 '12 at 11:02
  • So {0,{a},{b}} will also be a base i guess? – Haritz Nov 20 '12 at 11:05
  • @Haritz: Yes, it is; thanks for catching that. – Brian M. Scott Nov 20 '12 at 11:12
  • You are welcome @Brian M. Scott – Haritz Nov 20 '12 at 11:14
  • @Haritz: It's worth noting that if you're given a topology, then you've already got a basis. Any topology is necessarily a base for itself, so that task is simpler than you were making it. Now, for $X={a,b,c}$ (any finite set, $X$, really) and any given topology on $X$, there does exist a minimal base. In your example $\mathcal T=\bigl{\emptyset,{a},{b},{a,b}\bigr}$, we have $\mathcal B=\bigl{{a},{b},X\bigr}$ as a minimal basis for $\mathcal T$--that is, if we remove any of the three sets from $\mathcal B$, then the new collection is no longer a basis for $\mathcal T$. – Cameron Buie Nov 20 '12 at 15:10
  • "A subset $\mathscr{B}$ of $\mathscr{T}$ is a base for $\mathscr{T}$ if and only if every member of $\mathscr{T}$ (i.e., every open set in the space $\langle X,\mathscr{T}\rangle$) is a union of members of $\mathscr{B}$." Why is this iff? I would say only if. For e.g., a collection that is constructed from a base for $\mathscr{T}$ but with an extra set added would still satisfy that condition but will not generate $\mathscr{T}$, but a finer topology, I think. – guillefix Aug 06 '14 at 17:57
  • If $\mathcal{B}=\big{{a},{b},{a,b}\big}$ is a basis for a topology on ${a,b}$, so is it true that $\left { \bigcup_{\mathcal{C}\subseteq \mathcal{B}}\mathcal{C} \right }=\big{{a},{b},{a,b}\big}$? How can this be a topology on ${a,b}$ if it does not contain $\emptyset$? – UnknownW Nov 26 '15 at 03:17
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    @AjmalW: $\left{\bigcup\mathscr{C}:\mathscr{C}\subseteq\mathscr{B}\right}= \big{\varnothing,{a},{b},{a.b}\big}$; you forgot $\mathscr{C}=\varnothing$, whose union is $\varnothing$. – Brian M. Scott Nov 26 '15 at 03:20
  • @AjmalW: You’re welcome. – Brian M. Scott Nov 26 '15 at 03:26
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A base is a collection $B$ of sets such that every set in the topology can be written as a union of sets in $B$.

1) Can you write every set in $T$ as a union of $\{a\}$ and/or $\{b\}$?

2) Produce all the unions then you'll see what the topology is.

Rudy the Reindeer
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Let $(X,\tau)$ be a topological space and $\mathcal{B}\subseteq 2^X$. $$\mathcal{B} \text{ is a base for topology } \tau \text { on } X$$ $$:\Leftrightarrow$$ $$1) \mbox{ } \mathcal{B}\subseteq \tau$$ $$2) \mbox{ } (\forall A\in\tau)(\exists\mathcal{A}\subseteq\mathcal{B})(A=\cup\mathcal{A})$$

murad.ozkoc
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