Graph of a discontinuous function is closed

Question

I’m currently reading an book to introduce me to topology, and one of the exercises was to prove that the graph of the function $f: \mathbb{R} \rightarrow \mathbb{R}$ defined by $f(x)=0 \forall x \leq 0$ and $ f(x)=\frac{1}{x} \forall x>0 $ is closed. If the graph is closed, then the limit of any sequence of points will be a point on the graph. Since f is continuous away from 0, it’s clear this property will hold there. But if we take a sequence, say, $x_n=\frac{1}{n}$ which converges to 0 as n approaches infinity, and since $f(x_n)=n$ for any natural number n, we have $\lim_n (x_n,f(x_n))=(0, \infty)$. But this point isn’t in the graph, it’s not even in $\mathbb{R}^2$. I’m fairly certain that this point not being in $\mathbb{R}^2$ is what allows this set to be closed even though this is a counterexample of the limit property of closed sets. Is this correct? If so, why do we need points a sequence converges to to be in the space at hand ($\mathbb{R}^2$ in this case)? If we were to take $f: \mathbb{\overline R} \rightarrow \mathbb{\overline R}$ would this set not be closed for the reason I described above?

Answer 1

If you want to use sequences, what you have to prove is (denoting the graph by $\Gamma(f)$):

If we have $(x_n, y_n) \in \Gamma(f)$ such that $(x_n, y_n) \to (x,y)$ in $\Bbb R^2$ (or $\overline{\Bbb R}^2$ if you want to work in that space) then $(x,y) \in \Gamma(f)$

Now, if we're working in the "extended plane", then your argument would show $\Gamma(f)$ is not closed there: $(\frac1n, n) \in \Gamma(f)$ and $(\frac1n, n) \to (0,+\infty) \notin \Gamma(f)$. In fact for compact spaces (like $\overline{Bbb R}$) closedness of graph is equivalent to continuity)

But this does not mean that $\Gamma(f)$ is not closed in $\Bbb R^2$. In fact it is closed: suppose $(x_n, y_n)$ converges to $(x,y)$ in $\Bbb R^2$. Then we can distinguish several cases: $x <0$ which means that all but finitely many $x_n <0$ too and so we have a tail-sequence of $(x_n,0)$ which converges to $(x,0)\in \Gamma(f)$, and if $x>0$ we have that all but finitely many $x_n$ are $>0$ too and we have a regular sequence $(x_n, \frac{1}{x_n})$ that can only converge to $(x, \frac{1}{x}) \in \Gamma(f)$ by continuity of $\frac{1}{x}$ on $(0, \infty)$. Finally, $x=0$ is impossible for a convergent $(x_n, y_n) \to (x,y)$, (so $x_n \to 0$ and $y_n \to y$) as this would force $y_n= \frac{1}{x_n}$ to be unbounded, while we have $y_n \to y$ forcing $\{y_n: n \in \Bbb N\}$ to be bounded, contradiction. The criterion has been shown for $\Bbb R^2$ so $\Gamma(f)$ is closed in that set. Note that in both cases we can use that $(x_n, y_n) \to (x,y) \iff (x_n \to x) \land (y_n \to y)$ by the product topology/metric we use on the product.