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I'll put the following definitions here for completeness

Definition (Topology): A topology on a set $X$ is a collection $\mathcal{T}$ of subsets of $X$, having the following properties

  1. $\emptyset \in \mathcal{T}$ and $x \in \mathcal{T}$
  2. The union of the elements of any subcollection of $\mathcal{T}$ is in $\mathcal{T}$
  3. The intersection of the elements of any finite subcollection of $\mathcal{T}$ is in $\mathcal{T}$

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Definition (Basis for a Topology): If $X$ is a set, a basis for a topology on $X$ is a collection $\mathcal{B}$ of subsets of $X$ such that

  1. $\forall x \in X$ there exists at least one $B \in\mathcal{B}$ such that $x \in B$
  2. If $x \in B_1 \cap B_2$ (the intersection of two basis elements), then there exists a $B_3$, such that $x \in B_3$ and $B_3 \subset B_1 \cap B_2$

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Definition (Topology generated by a basis): If $X$ is any set, $(U \subset X) \in \mathcal{T}$ if $\forall x \in U$, there exits a basis element $B \in \mathcal{B}$ such that $x \in B$ and $B \subset U$


Right now, onto my question. If we define a basis for a topology on a set $X$ as follows $$\mathcal{B} = \{B_i\} \text{ where } \bigcup_i B_i = X \text{ and } B_{i} \cap B_j = \emptyset \text{ for } i \neq j$$

$\mathcal{B}$ satisfies properties 1 and 2 in the definition of a basis. That it satisfies property 1 is obvious and for property 2 it satisfies it vacuously as $\not\exists \ x$ such that $x \in B_1 \cap B_2$.

And while $X \not\in \mathcal{B}$, we have $X \in \mathcal{T}$ as $X \subset X$ and by our construction of $\mathcal{B}$, $\forall x\in X$, there exists a basis element $B_i \in \mathcal{B}$, such that $x \in B_i$ and $B_i \subset X$

Therefore this leads me to believe that a basis for a topology on a set $X$, need not contain $X$ itself. Am I correct?

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  • $\begingroup$ You are correct. Part 1. of the def'n only requires that every member of $X$ belongs to at least one element of $B.$ More succintly, $\cup B=X$. $\endgroup$ – DanielWainfleet Dec 28 '16 at 10:11
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Indeed. Since $\mathcal B$ is a basis, for any $x\in X$ there is some $B_x\in \mathcal B$ such that $x\in B_x$. Then $$X=\bigcup_{x\in X}B_x$$ So the basis generates $X$.

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