Easy topology inquiry 
(DEF 1): A topological space $X$ is a set with a collection $B$ of
  neighborhoods $N \subseteq X$ such that 
  
  
*
  
*for all $x \in X$, there exists $N \subseteq B$ such that $x \in N$
  
*for all $N_1, N_2 \in B$ with $x \in N_1 \cap N_2$, we have an $N_3 \in B$ such that $x \in N_3 \subseteq N_1 \cap N_2 $
$B$ is called a basis for the topology on $X$
(DEF 2): Let $X$ be a metric space. The metric topology on $X$ is
  defined by using set $N$ as neighborhoods of a point $x \in X$ where
$$D_X(x,r) = \{ y \in X : d(x,y) < r \} $$
where $r > 0$

(PROBLEM): Let $X$ be a metric space. Show that the collection of neighborhoods defined in (DEF 2) forms a basis for the metric topology.
SOLUTION: Take $x \in X$, and put $N = D_X(x,r)$. Then $x \in N$ since $d(x,x) = 0 < r$. So, condition (1) of def1 is satisfied. Now, put $N_1 = D_X(x_1,r_1)$ and $N_2 = D_X(x_2,r_2)$. Pick a point $x \in N_1 \cap N_2$. Therefore, we have $d(x_1, x) < r_1$ and $d(x_2,x) < r_2$. Now, put $N_3 = D_X(x_3, r_3)$ and let $r_3 = \min(r_1,r_2)$. Hence, any point in $N_3$ must satisfy $d(x_3,x) < \min (r_1,r_2) \leq r_1$. In other words, $x$ must lie in $N_3$ and $N_3 \subseteq N_1 \cap N_2 $. So, condition (2) of def1 is satisfied and the problem is solved.
Is this correct? I do not know If im doing what the problem is asking for. Can you give me some feedback? thanks a lot!
 A: Here is the picture I was talking about. Take it as a hint:

A: Some HINTS:


*

*It should be $N \in B$, not $N \subset B$. The basis $B$ is a collection of subsets of $X$.

*You can' define an $N$ the way you did before fixing some $r>0$. You should say, "let $r>0$ be any positive real number and $N=D_x(x,r)$ then...".

*For the second property you can already see an answer with a nice picture. Don't say "put $N_1$ [...] and $N_2$...". They are given. What you have to say is: "Let $N_1$ and $N_2$ be any two given members of $B$ containing $x$, now put (define) $N_3$....". 

*To see the radius that $N_3$ should have see the picture you have as an answer.

*To prove $N_3 \subset  N_1 \cap N_2$ you have to take an element other than  the center of $N_3$ and then by means of the triangle inequality show  that it is in $N_1 \cap N_2$. Once for $N_1$ and once for $N_2$ you will have to first show that this point is close enough to the center of the ball and then with the triangle inequality make use of the convenient radius you defined.

A: part (1) of definition (1): Put $N = D_X(x,r)$. Therefore $x \in N$ since $d(x,x) = 0 < r $.
part (2) of definition (2): Given $N_1 = D_X(x_1,r_1)$ and $N_2 = D_X(x_2,r_2)$ such that $x \in N_1 \cap N_2 $, we need to find $N_3$ such that $x \in N_3 \subseteq N_1 \cap N_2 $. Put $N_3 = D_X (x_3, r_3)$ where $r_3 = \min \{r_1 - d(x_1,x_3), r_2 - d(x_2,x_3)\}$. Now pick $y \in N_3$. Hence, $d(x_3,y) < r_3$. Now, 
$$ d(y, x_1) \leq d(y,x_3) + d(x_3,x_1) < \min \{r_1 - d(x_1,x_3), r_2 - d(x_2,x_3)\} + d(x_3,x_1) \leq r_1 - d(x_3,x_1) + d(x_3,x_1) = r_1$$
Therefore, $y \in N_1$. Similarly,
$$ d(y,x_2) \leq  d(y,x_3) + d(x_3,x_2) \min \{r_1 - d(x_1,x_3), r_2 - d(x_2,x_3)\} + d(x_3,x_2) \leq r_2 - d(x_2,x_3) + d(x_3,x_2) = r_2$$
Therefore, $y \in N_2$. In other words, we have shown that $N_3 \subseteq N_1 \cap N_2 $. However, I dont know how to show that actually $x \in N_3$. Can we just put $x_3 = x$ ??
