proving equality of basis in topological spaces The subset $Y=[0,1]\cup \{2\}$ of $\mathbf R$ is equipped with the topology $\tau$ generated by the basis $\mathscr B$ made of the following sets:


*

*$\{y \in Y\mid a<y<b\}$ for $a,b \in Y$ such that $a<b$.

*$\{y \in Y\mid 0\le y<b\}$ for $b \in Y$ 

*$\{y \in Y\mid a<y\le 2\}$ for $a \in Y$ 


Is $\tau$ equal to the subspace topology on $Y$? justify
Basically I started stating that the subspace topology is equal to the intersection $Y$ with something open in $\mathbf R$ but how can i compare them since this type of basis seems weird for me can anyone help 
Thank you..
 A: We want to decide wether two topologies $\tau_1,\tau_2$ on the same set $Y$ are equal. This is equivalent to showing that their basis coincide/do not coincide. As we are given the basis $\mathcal{B}_1$ on $Y=[0,1] \cup \{2\}$ we may compare it to the basis $\mathcal{B}_2$ that induces the subspace topology on $Y \subset \mathbb R$.
How does $\mathcal{B}_2$ look like?
We know that the set of open intervals $\mathcal{B} := \{(a,b) \mid a,b \in \mathbb R, a \neq b \}$ is a basis of the standard topology on $\mathbb R$. We deduce that: $$\mathcal{B}_2=\{(a,b) \cap Y \mid a,b\in \mathbb R, a \neq b \}$$
From here on split the problem in two steps:
1st step: $\mathcal{B}_1 \subset \mathcal{B}_2$
The basis element $Y \cap (a,b)$ (for $a,b \in Y$) lies in $\mathcal{B}_2$ as $Y \subset \mathbb R$. 
For the second set observe that if we intersect $(-3,b)$, which is an open interval in $\mathbb R$, with $Y$ we obtain $(-3,b) \cap Y= \{y \in Y \mid 0 \leq y <b\}$ and so it also lies in $\mathcal{B}_2$. The third one works in a similiar fashion (you 'jump' just to the right far enough with an open interval and intersect it with $Y$!)
2nd step: $\mathcal{B}_2 \subset \mathcal{B}_1$
This is basically again the very same argmuent as in the first step, however this does not happen in general that the two inclusions work the same. In this example it happens to be the case because it's easy to write down very explicit formulas for the sets. I just wanted to break this proof into two steps in order to show you how it's down systematicaly as this might be quite handy when working with some nasty topologies/basis.
A: Denote the subspace topology by $\tau_s$. 
We observe:


*

*$Y\cap(a,b)\in\tau_s$ for $a,b\in Y$

*$Y\cap[0,b)=Y\cap(-1,b)\in\tau_s$ for $b\in Y$

*$Y\cap(a,2]=Y\cap(a,3)\in\tau_s$ for $a\in Y$


We conclude that $\mathscr B\subseteq\tau_s$ and consequently $\tau\subseteq\tau_s$.
In order to prove that also $\tau_s\subseteq\tau$ let $r\in\mathbb R$. It is enough to prove that $Y\cap(r,\infty)\in\tau$ and $Y\cap(-\infty,r)\in\tau$ (do you understand why?). 
This is evidently true for $r<0$ and $r>2$ so focus on $r\in[0,2]$.
I leave the rest to you. Let me know if you get stuck again.
