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I came to know about 2 definitions of a subbasis of a topology. 1. A collection of subsets of X viz. S is said to be a subbasis of a topology T on X if S covers X and the finite intersections of elements of the subbasis form the basis elements (H.L.Royden). 2. If we consider the all the topologies on X containing S, then their intersection is also a topology on X and it is the smallest topology containing S. Then S is said to be a subbasis of the smallest topology, and the topology is said to be generated by S.

Now I want to prove their equivalency, i.e. S is a subbasis of a topology T iff T is generated by S. I have tried in the way: First let T is generated by S. Then by 2nd definition T is a topology containing S. Let B be the set of all finite intersections of elements of S.Then S is a subset of B and B is a subset of T (as T is a topology). LET T1 be the collection of possible unions of elements of B, we will show that T = T1. After this I am not able to proceed. Also I am unable to sketch the converse part. If something is wrong, please let me know, and also I hope to get help. Thank You.

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Let me rewrite your definitions;

1) $S$ is said to be a subbase of a topological space $(X,\mathbb{\tau})$ iff a) $\bigcup S=X$ b) If $\mathcal{B}$ consists of all finite intersections of elements of S then $\mathcal{B}$ is a base of $\tau$.

In other words for all $U\in\tau$ if $x\in U$ then there is a positive integer $n$ such that $U_1,U_2,\ldots U_n\in S$ with $x\in U_1\cap\ldots \cap U_n\subset U$

2)Let say $(X,\mathbb{\tau})$ a toplogical space. Then we say $\tau$ is generated by $\mathcal{S}$ iff $\tau=\bigcap \{\tau_i: \tau_i$ toplogy on $X$ and $S\subset \tau_i $} (Right here we have to add this "and $\bigcup S=X$". )

(If we do not put $\bigcup S=X$ in 2), we will have some trouble. For example say $S=\emptyset$ and so $\tau=\{X,\emptyset\}$ and the set of finite intersections of subsets of $S$ is not base for $\tau$.)

And you want to see 1) iff 2).

$1)\Rightarrow2)$

Since $S\subset \tau$ then $\bigcap \{\tau_i: \tau_i$ toplogy on $X$ and $S\subset \tau_i $} $\subset \tau$.

Take any nonemypty $U \in \tau$ and take any $x\in U$. Since we have $x\in U_1\cap\ldots \cap U_n\subset U$ and $U_1\cap\ldots \cap U_n \in \tau_i$ for all $\tau_i$ then we can say $x$ is an interior point of $U$ according for all $\tau_i$. So $U$ is open according for all $\tau_i$. Then $U\in \tau_i$ and so $\tau \subset \tau_İ$. Therefore $\tau\subset\bigcap \{\tau_i: \tau_i$ toplogy on $X$ and $S\subset \tau_1 $}.

$2)\Rightarrow1)$ It is clear that $S\subset \tau$ and $\tau$ is a toplogy. Say $\mathcal{B}$ consist of all fininite intersection of $S$ and $\tau_1$ be the collection of possible unions of elements of $\mathcal{B}$. And $\tau_1$ is a toplogy and since $S\subset \tau_1$ we have $\tau\subset \tau_1$. Take any nonemypty $U \in \tau_1$ and take any $x\in U$. Since we have $x\in U_1\cap\ldots \cap U_n\subset U$ and $U_1\cap\ldots \cap U_n \in \tau$, we can say $x$ is an interior point of $U $ according $\tau$. Then $U$ is open according $\tau$. In other words $U\in\tau$ and so $\tau_1=\tau$.

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  • $\begingroup$ Thanks a lot..Your answer and explanation cleared all my doubts.. $\endgroup$ – hiren_garai Apr 30 '17 at 0:40

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