2
$\begingroup$

I know that similar questions were already asked here, here and here. Unfortunately I wasn't able to understand the equivalence of definitions from any of answers (to any of the linked questions). I am new to topology and most of the answers to the linked questions felt very "hand wavy" and imprecise to me (even thought I might be missing something). Long story short, I’ve really been struggling to understand these answers, which is why I’m asking even though similar questions have been posted.

Definition - Let $\left(X, \tau \right)$ be a topological space. A collection of subsets $\mathcal{B} \subseteq \mathcal{T}$ is said to be a basis of $\mathcal{T}$ if it satisfies:

  1. For every $x\in X$ there exists $B\in \mathcal{B}$ such that $x\in B$.
  2. If for some $x\in X$ there exists sets $B_{1},B_{2} \in \mathcal{B} $ such that $x\in B_{i}$ ($i=1,2$) then there exists $B_{3} \in \mathcal{B}$ such that $x\in B_{3}$ and $B_{3} \subseteq \left(B_{1} \cap B_{2} \right)$.

I am trying to prove that

$\mathcal{B} \subseteq \mathcal{T}$ is a basis if and only if for every $U \in \mathcal{T}$ and for every $x\in U$ there exists $B\in \mathcal{B}$ such that $x\in B$ and $B \subseteq U$.

I got stuck trying to prove this implication $\Longrightarrow$.

Here is what I have tried:
Let $U \in \mathcal{T}$ and let $x\in U$, since $\mathcal{B}$ is a basis for $\mathcal{T}$ then by the first property there exists $B\in \mathcal{B}$ such that $x\in B$. $U$ satisfies $U = \bigcup_{u\in U} \left\{u \right\}$ and again by property $1$ for each $u \in U$ there is some $B_{u} \in \mathcal{B}$ such that $u\in B_{u}$, hence we get that $ U = \bigcup_{u\in U} \left\{u \right\} \subseteq \bigcup_{u\in U} B_{u}$. Since $x\in B , U$ we get that $x \in \left[ B \cap U \right] \subseteq \left[B \cap \left( \bigcup_{u\in U} B_{u} \right) \right] = \bigcup_{u\in U} \left( B \cap B_{u} \right)$. From here there exists $u\in U$ such that $x \in B \cap B_{u}$, by property $2$ there is some basis set $O \in \mathcal{B}$ such that $x\in O$ and $O \subseteq B \cap B_{u}$. This is where I get stuck (I don't know if $O$ has to be a subset of $U$ but I am not able prove it even if it is), every attempt I make to try and construct some basis set that contains $x$ and is contained in $U$ fails.

There is a repeating argument that I saw in many answers to similar questions (part of the linked questions as well) which I don't understand and was not mentioned during my lecture, that is;
arguing that $\mathcal{T}$ is the topology generated by $\mathcal{B}$ (which frankly I don't know the meaning of) and therefore there exists $\left\{ B_{i} \right\}_{i\in I} \subseteq \mathcal{B}$ such that $U = \bigcup_{i\in I} B_{i}$. I have even tried proving that this condition if equivalent to the definition of a basis as mentioned above, but I wasn't able to show that. Is there a way to prove this equivalence using only set theoretic manipulations (as well as the axioms)? (like was trying to do)

Thanks a lot!

$\endgroup$
6
  • 1
    $\begingroup$ The definition of a basis you quoted is incorrect. The family $\mathcal{B} = \{X\}$ satisfies the conditions whatever $\mathcal{T}$, but rarely is a basis. $\endgroup$ Commented Oct 18 at 9:39
  • $\begingroup$ @DermotCraddock thank you. Can you tell what is incorrect about it? it seems to me just like (one of the) definition in wikipedia $\endgroup$ Commented Oct 18 at 9:43
  • 1
    $\begingroup$ It doesn't require that $\mathcal{B}$ contains "small" open sets, which a basis must. However, what it correctly characterises are families that are a basis of some topology. If $\mathcal{B}$ is a family with the properties 1. and 2., then the family of all arbitrary unions of members of $\mathcal{B}$ is a topology $\mathcal{T_B}$, and $\mathcal{B}$ is a basis of $\mathcal{T_B}$. $\endgroup$ Commented Oct 18 at 9:48
  • $\begingroup$ @DermotCraddock So you are suggesting that $\mathcal{B} \subseteq \mathcal{T}$ that satisfies $1,2$ is a basis for some topology, that is not necessarily $\mathcal{T}$ (but might be contained in it)? $\endgroup$ Commented Oct 18 at 9:52
  • $\begingroup$ So when we say that $\mathcal{B}$ is a basis of $\mathcal{T}$ what we really mean is that $\mathcal{T} = \left\{ \bigcup_{i\in I} B_{i} : B_{i} \in \mathcal{B} \right\}$? $\endgroup$ Commented Oct 18 at 9:53

1 Answer 1

Reset to default
5
$\begingroup$

You cannot prove $\implies$ because your definition does not make sense. Indeed, there is no usable relationship between the topology $\mathcal T$ and the subset $\mathcal B$. For example, the set $\mathcal B = \{ X \}$ would be a basis for each topology $\mathcal T$ in the sense of definition, but you cannot use $\mathcal B$ to build any non-trivial element of $\mathcal T$.

The correct definition is this:

Definition 1. Let $X$ be a set. A collection $\mathcal{B}$ of of subsets of $X$ is said to be a basis for a topology on $X$ if it satisfies

  1. For every $x\in X$ there exists $B\in \mathcal{B}$ such that $x\in B$.
  2. If for some $x\in X$ there exists sets $B_{1},B_{2} \in \mathcal{B} $ such that $x\in B_{i}$ ($i=1,2$) then there exists $B_{3} \in \mathcal{B}$ such that $x\in B_{3}$ and $B_{3} \subseteq \left(B_{1} \cap B_{2} \right)$.

Note that $X$ does not yet have a topology, but the set $\mathcal B$ generates a topology $\mathcal T(\mathcal B)$ on $X$. Explicitly $$\mathcal T(\mathcal B) = \{ U \subset X \mid \text{ For all } x \in U \text{ there exists } B \in \mathcal B \text{ such that } x \in B \subset U \}.$$

It is easy to verify that $\mathcal T(\mathcal B)$ is a topology on $X$; actually it is the coarsest topology containing $\mathcal B$.

Your theorem is actually another definition:

Definition 2. Let $(X, \mathcal T)$ be a topological space. A subset $\mathcal{B} \subset \mathcal{T}$ is called a basis for the topology $\mathcal T$ if for every $U \in \mathcal{T}$ and for every $x\in U$ there exists $B\in \mathcal{B}$ such that $x\in B \subset U$.

The difference to Definition 1 is that the topology $\mathcal T$ is already given here.

Let us finally show that the any set $\mathcal B$ as in Definition 2 satisfies conditions 1. and 2. in Definition 1 (which is the reason why one defines an "abstract" basis on a set $X$ as in Definition 1).

  1. is trivial.

  2. Since $\mathcal T$ is a topology and $\mathcal{B} \subset \mathcal{T}$, the intersection $B_1 \cap B_2$ of any two $B_1, B_2 \in \mathcal{B}$ is an element of $\mathcal T$. It need not be an element of $\mathcal B$, but by definition for each $x \in U := B_1 \cap B_2$ there exist $B_3 \in \mathcal B$ such that $x \in B_3 \subset U$.

Also see my answer to Does a basis exist independent of the topology?

$\endgroup$
3
  • $\begingroup$ I find it very weird that we define the notion of a basis to a topology without including a topology in the definition. Thank you for that answer! $\endgroup$ Commented Oct 18 at 14:13
  • $\begingroup$ Sorry for all the comments but I just want to make sure that I understand it correctly. Given a topological space $\left( X, \tau \right)$ we say that a collection of subsets $\mathcal{B} \subseteq \mathcal{P} (X)$ is a basis for $\tau$ if $\tau = \left\{ \bigcup_{i \in I} B_{i} : B_{i}\in \mathcal{B} \right\}$. And regardless given a set $X$ (without an explicit topology) we say that a collection of subsets $\mathcal{B}$ is a basis for a topology if it satisfies theses two conditions. is this correct? $\endgroup$ Commented Oct 18 at 14:49
  • 1
    $\begingroup$ @Groot_loves_math Correct! $\endgroup$ Commented Oct 18 at 15:32

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.