Example of a closed map which is not continuous.

Question

A map $F: X \to Y$ is said to be closed if $x_n \to x$ in $X$ and $F(x_n) \to y $ in $Y$ implies $y = F(x)$.

As confusing as the definition is, I cannot seem to understand how a closed map is not necessarily continuous. There is also an attached example:

Let $F: \mathbb{R} \to \mathbb{R}$ be given as $F(t) = 1/t$ if $t \ne 0$, and $F(0) =0$. Then $F$ is a closed map but not continuous. How?

I tried taking the sequence $t_n = 1/n$ which converges to $0$, but $F(x_n) = n$ which doesn't converge so the definition doesn't hold. I understand that I need a sequence $x_n$ such that $x_n \to 0$ but $F(x_n) \to y \ne 0 = F(0)$, but I cannot think of such a sequence.

Answer 1

• "So the definition doesn't hold" is unclear. I would rather say "so $F$ is not continuous at $0$".
• "I need a sequence $x_n$ such that $x_n \to 0$ but $F(x_n) \to y \ne 0 = F(0)$": no, you don't. $F$ is closed because:
  • If a sequence $x_n\to x$ is eventually constant and $F(x_n)\to y\in\Bbb R,$ then $F(x)=y$ (by uniqueness of the limit). So, we only have to check the condition for sequences which are not eventually constant.
  • For $x=0$ the condition "for every non-eventually-constant sequence $x_n\to x$ s.t. $F(x_n)\to y \in\Bbb R $ we have $F(x)=y$" is vacuously true, precisely because there exists no such sequence (prove it).
  • For every $x\ne0$ the condition is also true, by continuity of $F$ at such a point.