Writing down a basis for usual topology on $\mathbb{R}$ on $[a,b)$ For $\mathbb{R}$ we have $\mathcal{B}=\left\lbrace (a,b):a<b, a,b \in \mathbb{R} \right\rbrace$. And let's denote basis for $[a,b)$ by $\mathcal{B}_{[a,b)}$. Then we have:
$$\mathcal{B}_{[a,b)} = \left\lbrace [a,b)\cap(a,b):a<b, a,b \in \mathbb{R} \right\rbrace$$
Since $\forall a,b\in \mathbb{R}$ satisfies $(a,b)\subseteq[a,b).$ We can say, $(a,b)\cap[a,b)=(a,b)$. Therefore we conclude:
$$\mathcal{B}=\mathcal{B}_{[a,b)}$$
And by same reasoning we can say the other way:
$$\mathcal{B}=\mathcal{B}_{(a,b]}$$ and $$\mathcal{B}=\mathcal{B}_{[a,b]}$$
Is it true? Thanks in advance!
 A: I think your choice of notation has caused some confusion here. Fix $a,b\in\mathbb{R}$. The basis $\mathcal{B}$ consists of all open intervals of the form $(x,y)$ where $x<y$ are real numbers. Now
$$\mathcal{B}_{[a,b)} = \{[a,b)\cap (x,y): x,y\in\mathbb{R}, x<y\}.$$
From this we see there are a few cases. If $a\leq x<y\leq b$, then the intersection is just $(x,y)$. If $y>b$, then the intersection is $(x,b)$. If $x< a$, however, then the intersection is $[a,y)$ or $[a,b)$ depending on whether $y\leq b$ or $y>b$. So the basis $\mathcal{B}_{[a,b)}$ consists of the usual open intervals so long as they fit inside $[a,b)$ together with half-open intervals of the form $[a,y)$ for $y\leq b$. In particular $\mathcal{B}_{[a,b)}\neq \mathcal{B}$ since the half-open interval $[a,b)$ belongs to the former but not the latter.
Similar considerations apply to $\mathcal{B}_{(a,b]}$ and $\mathcal{B}_{[a,b]}$. In the first we have open intervals which fit completely inside $(a,b]$ and also rays of the form $(x,b]$. In the second we have open intervals which fit completely inside $[a,b]$, half-open intervals of the form $[a,y)$, half-open intervals of the form $(x,b]$, and also the entire space $[a,b]$.
