Continuity of addition under the lower limit topology Letting $\Bbb R_\ell$ denote $\mathbb{R}$ with the lower limit topology, prove that the function $f : \Bbb R_\ell \times \Bbb R_\ell \to \Bbb R_\ell$ defined by $f((x,y)) = x+y$ is continuous.
 A: Computational error corrected.
Start with a basic open set in the range, say $[a,b)$. Its inverse image under addition is
$$\{\langle x,y\rangle\in\Bbb R^2:a\le x+y<b\}\;;$$
geometrically this is the diagonal strip between the lines $x+y=a$ and $x+y=b$, including the former edge but not the latter. Take a point $p=\langle x_0,y_0\rangle$ on the line $x+y=a$. The distance from $p$ to the line $x+y=b$ is $b-a$, but that’s in the direction perpendicular to the strip, i.e., in the direction parallel to the line $y=x$. Thus, the point on the line $x+y=b$ closest to $p$ is the point
$$q=\left\langle x_0+\frac1{\sqrt2}(b-a),y_0+\frac1{\sqrt 2}(b-a)\right\rangle\;,$$
and $p$ and $q$ are diagonally opposite corners of the open box
$$\left[x_0,x_0+\frac1{\sqrt2}(b-a)\right)\times\left[y_0,y_0+\frac1{\sqrt2}(b-a)\right)$$
in the Sorgenfrey topology. Now just check that the strip is the union of all such open boxes.
A: The preimage of an open set $[a,b)$ is 
$$
\{(x,y)\colon a \le x+y < b\}.
$$
To prove that such a stripe is open in the product space, write it as the union of squares $[x,x+(b-a)/2) \times [a-x,a-x+(b-a)/2)$ for $x \in \mathbb R$.
