Check that set is basis of topology

Question

I have set $ X =\mathbb{R}$ and I should to check that set $B$ is basis of topology, where $U \in B \Leftrightarrow U = [x, x+1)$

I started to check conditions, but I stuck on second condition and I dont know what to do next. I show how I did it:

1) $\exists x \in X$, that $U \in B$, where $x \in U$. Its clear, because $x \in [x,x+1)$. This condition holds.

2) if $U_{1} \in B, U_{2} \in B$, then for $x \in U_{1} \cap U_{2}$ exists set $V \in B$, where $ x \in V$. I started to do it like this $U_{1} = [x,x+1),\, U_{2} = [y,y+1]$ and $U_{1} \cap U_{2} = [\max{(x,y)}, \min{(x+1, y+1)})$.
If x < y, we can get $U_{1} \cap U_{2} = [y, x+1)$....

Here I stuck and dont know what to do with that.

Answer 1

The second condition is more stringent.

If $U_1,U_2\in B$ and $x\in U_1\cap U_2$ there exists $V\in B$ such that

• $x\in V$

• $V\subseteq U_1\cap U_2$

In particular the intersection, if not empty, must contain an interval of length $1$ without the upper end. With $U_1=[0,1)$ and $U_2=[-1/2,1/2)$, the intersection is $$ U_1\cap U_2=[0,1/2) $$ that cannot contain intervals of the form $[t,t+1)$.

Answer 2

Since the second condition doesn't hold, you have proved that $B$ is not a basis of a topology.