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Let $X$ be a set. Suppose $\beta$ is a basis for the topology $\tau_\beta$ of $X$. Since each base element is open (with respect to $\tau_\beta$) we have that $$B\in \beta\Rightarrow B\in \tau_\beta.$$ Thus, $\beta\subset \tau_\beta$.

However, since $\beta$ is a union of base elements (I assume a set can always be written as a union of its elements) and topologies are closed under arbitrary unions, we have $\beta \in \tau_\beta.$ Is it possible for a set to be a subset an an element of another set?

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Your assumption that a set can always be written as a union of its elements is completely wrong. Consider for example $\{ \varnothing\}$, where $\varnothing$ is the empty set. – Chris Eagle Dec 11 '12 at 23:41
The elements of $\tau_\beta$ are subsets of $X$. $\beta$ is not a subset of $X$, the elements of $\beta$ are subsets of $X$. – copper.hat Dec 11 '12 at 23:45
up vote 3 down vote accepted

Let $S$ be any set. Then $$S=\bigcup_{x\in S}\{x\}\;,$$ but in general $$S\ne\bigcup_{x\in S}x\;.$$ That is, $S$ is the union of the singletons of its elements, but it is not in general the union of its elements. It can only be the union of subsets, and in general $x\in S$ does not imply that $x\subseteq S$. (Sets $S$ with the property that $x\subseteq S$ whenever $x\in S$ are called transitive and are very important in set theory, but they’re the exception, not the rule.)

In particular, your $\beta=\bigcup_{B\in\beta}\{B\}=\{B:B\in\beta\}$; $\bigcup_{B\in\beta}B$ is a completely different thing. In fact we know that $\bigcup_{B\in\beta}B=X$, because one of the conditions defining a base for a topology on $X$ is that every point of $x$ belong to some member of the base.

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$\beta$ is a set of sets, and is not the union of it's elements .$\beta$ is not a union of basis elements, but a union of sets containing basis elements. Consider for example $A = \{[0,1], [1,2]\}$. It should be clear that $A \not= [0,2]$.

However, in general it is possible for a set to be both a subset and an element of another. Take for example $A = \{1\}, B =\{1,\{1\}\}$. Certainly $A \subset B$ and $A \in B$.

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