Open and Closed Quotient Maps. 
a) Find a subset $A$ of $\Bbb{R}$ such that the quotient map $p: \Bbb{R} \rightarrow \Bbb{R}/A$ is not open.

If we let $A= \Bbb{Q}$, then we can see that $(0,1)$ is open in $\Bbb{R}$. But if we pick a rational number $p/q \in (0,1)$ and draw a ball around it in $\Bbb{R}/\Bbb{Q}$, we will see that $0$ and $1$ are not contained in the ball...they they are equal to $p/q$ since they are rational. This is a contradiction, right?

b) Find a subset $B$ of $\Bbb{R}$ such that the quotient map $q : \Bbb{R} \rightarrow \Bbb{R}/B$ is not closed. 

Let $A= \{a,b\}$ for $a,b \in \Bbb{R}$ such that $a < b$. Then $[a,b]$ is closed in $\Bbb{R}$, since we cannot have a ball around the point $a$ without including points outside of the interval. However, in $\Bbb{R}/\{a,b\}$, we can just take the ball with radius $\frac{b-a}{2}$. Since $b=a$ in $\Bbb{R}/\{a,b\}$, we don't have to include points outside that interval.

c) Using on of these examples, show that a quotient space of a Hausdorff space need not be T1. 

We can look at $\Bbb{R}/\Bbb{Q}$. If we take a rational number $q \in \Bbb{Q}$ and an irrational number $a \in \Bbb{R}$, we cannot draw a ball around $a$ without including $q$ since the rational numbers are dense in $\Bbb{R}$ and all rational numbers are equal in $\Bbb{R}/\Bbb{Q}$. 
Are my answers correct? If not, can you give me any hints?
Thanks in advance 
 A: It’s true that the quotient map from $\Bbb R$ to $\Bbb R/\Bbb Q$ is not open, but your argument doesn’t really make sense: $\Bbb R/\Bbb Q$ isn’t a metric space, so it’s not clear what you mean by a ball in $\Bbb R/\Bbb Q$, and $0$ and $1$ aren’t points of $\Bbb R/\Bbb Q$. Let $q$ be the quotient map; then $$q^{-1}\big[q[(0,1)]\big]=(0,1)\cup\Bbb Q\;,$$ which is not open in $\Bbb R$, so $q[(0,1)]$ is not open in $\Bbb R/\Bbb Q$, and $q$ is therefore not an open map.
The same quotient space works for (b). This time consider the closed set $F=[0,1]$; $$q^{-1}\big[q[F]\big]=[0,1]\cup\Bbb Q\;,$$ which is not closed in $\Bbb R$, so $q[F]$ is not closed in $\Bbb R/\Bbb Q$, and $q$ is not a closed map.
Your example for (b) does not work: if $f:\Bbb R\to\Bbb R/\{a,b\}$ is the quotient map, then $f$ is closed. Let $F$ be any closed set in $\Bbb R$. If $F\cap\{a,b\}=\varnothing$, then $f^{-1}\big[f[F]\big]=F$, so $f[F]$ is closed in $\Bbb R/\{a,b\}$. Otherwise, $f^{-1}\big[f[F]\big]=F\cup\{a,b\}$, which is closed in $\Bbb R$, so again $f[F]$ is closed in $\Bbb R/\{a,b\}$, and $f$ is a closed map.
Your answer to (c) is fine.
