Confusion about concept of basis in point set topology. I'm afraid that I have a big misunderstanding about the notion of basis in general topology. For a given topology $\tau$ of set $X$, if there is a collection $S \subset \tau$ of open subsets of $X$ satisfying


*

*For any $x \in X$, there exists $A$ in $S$ such that $x \in A$

*For any $A,B$ in $S$ and for any $x \in A \cap B$, there exists $C$ in $S$ such that $x \in C \subseteq A \cap B$


Are those two condition not sufficient to say that $S$ is a basis of the initially given topology $\tau$ ? (That is, collection of arbitrary union of elements of $S$ might be not equal to topology $\tau$?)
Thanks in advance.
 A: The conditions you wrote do indeed say that $S$ is a basis for some topology $\tau'$. They also guarantee that $\tau' \subset \tau$. But, they do not guarantee that $\tau = \tau'$; see the counterexample in the comment of @DanielFischer. What you are missing is the axiom which says for any open subset $U \subset X$ and any $x \in U$ there exists $A$ in $S$ such that $x \in A \subset U$.
A: The two conditions you give are sufficient to ensure that $S$ is a basis for some topology, but it won't necessarily be the topology you started with.
What the condition $S\subset \tau$ will ensure is that the topology that has $S$ as a basis will be coarser than (or equal to) $\tau$, but no more than that.
For example (as Daniel Fischer noted), no matter what $\tau$ is, $S=\{X\}$ will always satisfy the condition -- conversely, if $\tau$ is the discrete topology, $S$ could be a basis for any topology on $X$.
A: First off, here is the formal definition of being a basis for a topology (notice that I bolded the word "a" -- after you read my answer, if you don't understand why, please comment):
Let $X$ be a set.  Let $\mathcal{A}$ be a collection of subsets of $X$.  Then $\mathcal{A}$ forms a basis for a topology if:


*

*For each $x \in X$, there is some $A \in \mathcal{A}$ such that $x \in A$.

*If $y \in A_{1} \cap A_{2}$ for some $A_{1}, A_{2} \in \mathcal{A}$, then there is some $A_{3} \in \mathcal{A}$ such that $y \in A_{3} \subseteq A_{1} \cap A_{2}$.
Notice that nowhere in the definition above did we say that $X$ is already a topological space with a given topology.  We just said $X$ is a set.  Then $\mathcal{A}$ was just a collection of subsets of $X$.  If $\mathcal{A}$ satisfies the two conditions above, it forms a basis for a topology.  Which topology?  It's always the same topology: the collection of all possible unions of elements of $\mathcal{A}$.  You could call this topology $T_{\mathcal{A}}$.
Now, if you are already given a topology $\tau$ on $X$, and you are given a collection $\mathcal{A} \subseteq \tau$, and $\mathcal{A}$ satisfies the two conditions above, it is now a basis for $T_{\mathcal{A}}$, as we discussed above.  Also, since $\tau$ is closed under arbitrary unions, it's obvious $T_{\mathcal{A}} \subseteq \tau$.  But, as Daniel Fischer's comment shows, it's not always true that $\tau \subseteq T_{\mathcal{A}}$.
So, to show that $\mathcal{A}$ is a basis for $\tau$, in addition to showing the above two conditions, so that it forms a basis for $T_{\mathcal{A}}$ (the topology that is the set of all possible unions of elements of $\mathcal{A}$), you also need to show $\tau \subseteq T_{\mathcal{A}}$, i.e., that every open set in $\tau$ can be written as a union of elements of $\mathcal{A}$.
There is one lemma that is useful (and you should prove) which can help you prove a collection of subsets of $\tau$ actually forms a basis for $\tau$:
Lemma. If $(X, \tau)$ is a topological space and $\mathcal{A} \subseteq \tau$ satisfies 1. and 2. above, then $T_{\mathcal{A}} = \tau$ iff for each $U \in \tau$ and $x \in U$, there is some $A_{x} \in \mathcal{A}$ such that $x \in A_{x} \subseteq U$.
Once you prove the above lemma, to show $\mathcal{A}$ is a basis for a given topology $\tau$, you need to show 1. and 2. are satisfied (which would give $T_{\mathcal{A}} \subseteq \tau$), and then to show $\tau \subseteq T_{\mathcal{A}}$, you just need to prove if $U \in \tau$ and $x \in U$, there is some $A_{x} \in \mathcal{A}$ such that $x \in A_{x} \subseteq U$.
