I was reading Topology from Munkres and got confused by the definition of a subbasis. What is/are the difference between basis and subbasis in a topology?
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$\begingroup$Bases and subbases "generate" a topology in different ways. Every open set is a union of basis elements. Every open set is a union of finite intersections of subbasis elements.
For this reason, we can take a smaller set as our subbasis, and that sometimes makes proving things about the topology easier. We get to use a smaller set for our proof, but we pay for it; with a subbasis we need to worry about finite intersections, whereas we did not have to worry about that in the case of a basis.
$\endgroup$ 2 $\begingroup$Consider $S=\{\{0,1\},\{0,2\}\}$. What is the topological space $T(S)$ generated by $S?$ By definition, $S$ will then be a subbasis of $T(S)$.
Well, we want to all requirements to hold true and find that $T(S) = \{\emptyset, \{0\}, \{0,1\}, \{0,2\}, \{0,1,2\}\}$ (check this!).
Is $S$ a basis? No, because you cannot write $\{0\}$ as a union of any elements in $S$.
So you see that subbasis and basis are two different notions, even for a very basic example.
A subbasis can be thought of, and is actually defined to be, the "smallest set that becomes my topological space if I complete it under the property of being a topological space, i.e. fulfiling the axioms of topological space".
The two terms are related nevertheless. Every basis is a subbasis, and in one of the equivalent definitions of subbasis you will find that you already get a basis from your subbasis.
$\endgroup$ 6 $\begingroup$The collection of sets $(-\infty,b)$ and $(a,\infty)$ for $a,b\in \mathbb{R}$ constitute a sub-basis for the standard topology on $\mathbb{R}$. The collection of sets $(a,b)$ for $a,b\in \mathbb{R}$ constitute a basis for the standard topology on $\mathbb{R}$. I suggest that you look at the definitions of "basis" and "sub-basis" and convince yourself that the claims that I have made are correct; this is probably the best way to answer your question.
The basic idea is that a basis is the collection of all finite intersections of sub-basis elements. The open sets in a topology are all possible unions of basis elements. So, the open sets in a topology are all possible unions of finite intersections of sub-basis elements.
I hope that answers your question!
$\endgroup$ 1 $\begingroup$If we ignore, momentarily, the fact that we are trying to generate a topology, a subbasis is any old collection of subsets of the space. Thus, any basis is a subbasis. However, a basis $\mathcal B$ must satisfy the criterion that if $U,V\in\mathcal B$ and $x$ is an arbitrary point in both $U$ and $V$, then there is some $W$ belonging to $\mathcal B$ such that $x\in W\subseteq U\cap V$.
To really understand what that means you have to look at examples. The family $\{X\}$ is a basis for the indiscrete topology on $X$. The family $\mathcal P(X)$, the powerset of $X$, is a basis for the discrete topology on $X$. If $X$ is large enough and you remove some subset $S\subset X$ from $\mathcal P(X)$, you can still have a basis if $S$ is not singleton but $\mathcal P(X)\setminus \{x\}$ is merely a subbasis when $x\in X$.
A more ambitious example is the definition of the topology on the product of two spaces $X$ and $Y$. The family $$\{U\times Y:U\subseteq X, U\text{ is open}\}\cup\{X\times V:V\subseteq Y,V\text{ is open}\}$$ is a subbasis but not necessarily a basis of the box topology on $X\times Y$.
Finally, suppose a family $\mathcal B$ is a basis. You can generate a topology $\tau_0$ using $\mathcal B$ as a subbasis and you can generate a topology $\tau_1$ using $\mathcal B$ as a basis. The topologies are actually the same, as you can see here. You just have to do more work to generate a topology from a subbasis $\mathcal S$, i.e. you have to include all finite intersections of members of $\mathcal S$ because it is not guaranteed that they are unions of members of $\mathcal S$.
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