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: $d(x,y)=\mathrm{\mid }x-y\mathrm{\mid }$ 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 $d_{\mathrm{\infty }}$ 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 ${\mathrm{\Delta }}_X\subseteq X\times X$, the diagonal map ${\delta }_X$, and the pairing $⟨f,g⟩$ of two maps)

${\mathrm{\Delta }}_{\mathbb{R}}=\{(t,t):t\in \mathbb{R}\}$

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

$(a-r,a+r)\times (b-r,b+r),r:=\frac{1}{2}\mathrm{\mid }a-b\mathrm{\mid }>0.$

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 $\mathrm{\mid }a-b\mathrm{\mid }\mathrm{/}2$ 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 $d_{\mathrm{\infty }}$ 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.