The diagonal of $\mathbb{R}$ is closed in $\mathbb{R}^2$, computed from the product basis

Give $\mathbb{R}$ its usual topology (The absolute value makes $\mathbb{R}$ a metric space: $d(x,y) = |x-y|$ is a metric, its open balls are the intervals $(x-r, x+r)$, and it is unbounded, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not) and let $\mathbb{R}^2 = \mathbb{R} \times \mathbb{R}$ carry the product topology (The product set $\prod_{i \in I} X_i$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space, For $n \ge 1$ the product topology on $n$ copies of the usual topology of $\mathbb{R}$ is the metric topology of $d_\infty$ on $\mathbb{R}^n$, and hence also of $d_1$ and $d_2$, so $\mathbb{R}^n$ as a product and $\mathbb{R}^n$ as a metric space are one space). Then the diagonal (The diagonal $\Delta_X \subseteq X \times X$, the diagonal map $\delta_X$, and the pairing $\langle f, g \rangle$ of two maps)

$\Delta_{\mathbb{R}} \;=\; \{\, (t,t) : t \in \mathbb{R} \,\}$

is closed in $\mathbb{R}^2$, and the box that separates a point $(a,b) \notin \Delta_{\mathbb{R}}$ from it may be written down:

$(a - r, a + r) \times (b - r, b + r), \qquad r := \tfrac{1}{2}|a - b| > 0 .$

Nothing here appeals to the general criterion; the computation is carried out against the product basis directly. It agrees with A space is Hausdorff if and only if its diagonal is closed in the square carrying the product topology, $\mathbb{R}$ being Hausdorff (Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not), and the point of writing it out is to show what the criterion's abstract box is in this case: the two open intervals of half the distance between the coordinates.

• The radius $|a-b|/2$ is exactly the Hausdorff separation of $a$ and $b$ in $\mathbb{R}$, and that is not a coincidence. The proof of A space is Hausdorff if and only if its diagonal is closed in the square carrying the product topology builds its box out of a pair of disjoint open sets separating the two coordinates; here that pair is $(a-r, a+r)$ and $(b-r, b+r)$, the two balls of radius half the distance which the usual metric supplies.

• The product topology on $\mathbb{R}^2$ is the topology of the usual metrics on it, so the computation above may be read equally as a statement about boxes or about balls (For $n \ge 1$ the product topology on $n$ copies of the usual topology of $\mathbb{R}$ is the metric topology of $d_\infty$ on $\mathbb{R}^n$, and hence also of $d_1$ and $d_2$, so $\mathbb{R}^n$ as a product and $\mathbb{R}^n$ as a metric space are one space); the box form is used because it is what the criterion tests.