Sequences in Topologies Assume that in $\mathbb{R}$ with the usual topology $p_n\to p,\ q_n\to q$ and that $p\neq q.$ Construct a new topology where $q_n\to p$ but $p_n\not\to q.$
I thought the best method would be to have one of them be constant since eventually constant sequences converge I believe in any topology. But no luck. I tried also using the trick with the indiscrete topology but no luck again. Any hints would be greatly appreciated.
 A: Let $\mathcal{T} = \{\phi, \{q\}, \mathbb{R}\}$. You can check that $\mathcal{T}$ is a topology on $\mathbb{R}$.
We have $q_n \to p$ with respect to $\mathcal{T}$ because the only open neighbourhood of $p$ is $\mathbb{R}$, which contains all the elements $q_n$.
However, $p_n \not\to q$ with respect to $\mathcal{T}$:
Since $\{q\}$ is an open neighhbourhood of $q$, any sequence converging to $q$ must eventually be equal to $q$. But $(p_n)_{n=1}^\infty$ converges to $p \ne q$ in the usual topology, so it cannot eventually be equal to $q$.
A: For what it's worth, here's a solution which does not always work, as @G.Sassatelli points out (and I had to correct an earlier draft). Namely, it works as long as none of the $p_n$'s or $q_n$'s are ever equal to $p$ or $q$.
Pick a point 
$$r \in \mathbb{R} - \bigl(\{p,q\} \cup \{p_1,p_2,\ldots\} \cup \{q_1,q_2,\ldots\}\bigr)
$$
Such a point exists because you are removing a countable set from an uncountable set.
Let $f : \mathbb{R} \to \mathbb{R}$ be the bijection
$$f(x) = \begin{cases}
p &\quad\text{if $x=q$} \\
r &\quad \text{if $x=p$} \\
q &\quad\text{if $x=r$} \\
x &\quad\text{otherwise}
\end{cases}
$$
Define a topology on $\mathbb{R}$ for which $U$ is open if and only if $f(U)$ is open in the standard topology. Then $q_n \to p$ and $p_n \to r \ne q$.
