Why does $f = 1/x, x \neq 0$ have a sequentially closed graph? Wikipedia says:

Let $X$ and $Y$ both denote the real numbers $\mathbb{R}$ with the usual Euclidean topology. Let $f: X \rightarrow Y$ be defined by $f(0) = 0$ and $f(x) = 1/x$ otherwise. Then $f:X \rightarrow Y$ has a closed graph (and a sequentially closed graph) in $X \times Y = \mathbb{R}^2$ but it is ''not'' continuous.

I don't understand why $f$ has a sequentially closed graph. Consider the sequence $\{1/n\} \subset X$ -- the image of this sequence under $f$ diverges in $Y$. For a closed graph, shouldn't the image of all convergent sequences have a limit? It's likely I'm not understanding the definition of a sequentially closed graph properly.
 A: What's required by the definition of a sequentially closed set is that for any sequence of points $(x_n,f(x_n))$ on the graph which converges to some point $(a,b)$ in the surrounding space $\mathbf{R}^2$, the limit also belongs to the graph (i.e., $b=f(a)$).
Your sequence $(1/n,n)$ does not converge in $\mathbf{R}^2$, so it's not relevant here.
A: A function $f : X \to Y$ has a sequentially closed graph if, well, the graph
$$\Gamma(f) := \{(x, f(x)) \in X \times Y : x \in X\}$$
is a sequentially closed subset of $X \times Y$ (under the product topology).
What does it mean for a subset $A$ of a topological space $Z$ to be "sequentially closed"? It means that, whenever you have a convergent sequence $z_n \in Z$ that converges to a point $l \in A$, the point $l$ must also lie in $Z$. Importantly, we only draw conclusions when $z_n$ has a limit. The definition is not violated if we find a divergent sequence $z_n$.
Now, in this case, $\frac{1}{n}$ is does not belong to $\Gamma(f)$; the sequence you're thinking of is $(1/n, 1/(1/n)) = (1/n, n)$. Yes, this is a sequence in $\Gamma(f)$. But no, this is not a convergent sequence in $X \times Y$ under the product topology. If it were, and it converged to $(x, y) \in X \times Y$, then the product topology would tell us that $\frac{1}{n} \to x$ and $n \to y$. The latter, in particular, is absurd.
So, we have found a sequence in the graph that fails to converge, but this is OK. The definition of sequentially closed sets puts no restrictions on divergent sequences.
