Find a continuous function that is neither open nor closed. Here was my attempt. Can I get a solution verification?
Consider the map $f:\mathbb{R}_l \rightarrow \mathbb{R}$ given by $f(x)=x$. Then if $\tau'$ is the lower limit topology on $\mathbb{R}$ and $\tau$ is the standard topology on $\mathbb{R}$, since $\tau \subset \tau'$, if $U$ is open in $\mathbb{R}$ then $f^{-1}(U)=U$ is open in $\mathbb{R}_l$. However $[0,1) \in \tau'$, but $f([0,1))=[0,1)$ and $[0,1) \notin \tau$. So $f$ is not open. Similarly $(-\infty,0) \cup [1,\infty)$ is closed in $\mathbb{R}_l$ but $f((-\infty,0) \cup [1,\infty))=(-\infty,0) \cup [1,\infty)$ and $(-\infty,0) \cup [1,\infty)$ is neither open nor closed in $\mathbb{R}$.
Best Answer
That example is fine. You can more simply note that $[0,1)$ is open in $\tau'$ but not open in $\tau$ (because of $0$ not being an interior point) and also closed in $\tau'$ but not closed in $\tau$ (because of $1$ being an outside limit point). So one set works for both refuting openness and closedness of $f$.