Continuity of functions of topological spaces Let $\mathbb{R}_l$ be the lower limit topology on the real line generated by the basis $\{[a,b) : a<b\}$, and let $\mathbb{R}$ be the standard topology generated by the basis $\{(a,b) : a<b\}$.
In my rather poorly explained topology notes, in one of the examples it says the addition $\sigma\colon\mathbb{R}_l\times\mathbb{R}_l \to \mathbb{R}_l$ defined by $\sigma(x,y)=x+y$ is continuous, but not continuous as a function $\sigma\colon\mathbb{R}_l\times\mathbb{R}\to\mathbb{R}_l$. Now I get the part of explanation why we have to show that the set $\sigma ^{-1}\bigl([a,b)\bigr)$ is open, but dont know why it is open with respect to the basis  $\{[a,b) \times [c,d)\}$ and not open with respect to $\{[a,b) \times (c,d)\}$. 
Could anyone help me clarify why this is the case? 
 A: [Update] More detailed statements are added by OP's request.
It is easy if you use the following Theorem:

Definition. Let $f\colon X\to Y$ be a function. We call $f$ is continuous at $x\in X$ if for every open set $V\subset Y$ containing $f(x)$, there exists an open set $U\subset X$ containing $x$ such that $f(U)\subset V$.
Theorem. A function $f\colon X\to Y$ is continuous if and only if for every $x\in X$, $f$ is continuous at $x$.



*

*By the theorem above, to prove the continuity of the addition $\sigma\colon\mathbb{R}_l\times\mathbb{R}_l\to\mathbb{R}_l$, it suffices to show that for every $(a,b)\in\mathbb{R}_l\times\mathbb{R}_l$, $\sigma$ is continuous at $(a,b)$. Now for any open set $V\subset\mathbb{R}_l$ containing $\sigma(a,b)=a+b$, we need to find an open set $U\subset\mathbb{R}_l\times\mathbb{R}_l$ containing $(a,b)$ such that $\sigma(U)\subset V$. Since $V$ is an open set in $\mathbb{R}_l$ containing $a+b$, there exists $\epsilon>0$ such that (a basis element) $[a+b,(a+b)+\epsilon)\subset V$. We can choose an open set
$$
U=[a,a+\tfrac{\epsilon}{2})\times[b,b+\tfrac{\epsilon}{2})\subset\mathbb{R}_l\times\mathbb{R}_l
$$
containing $(a,b)$. Then $\sigma(U)=[a+b,(a+b)+\epsilon)\subset V$. $\quad\square$

*We will show that the addition $\sigma\colon\mathbb{R}_l\times\mathbb{R}\to\mathbb{R}_l$ is not continuous at any $(a,b)\in\mathbb{R}_l\times\mathbb{R}$. Fix a point $(a,b)$. By the theorem above, to prove the discontinuity of $\sigma$ at $(a,b)$, we need to show that there exists an open set $V\subset\mathbb{R}_l$ containing $\sigma(a,b)=a+b$ such that for any open set $U\subset\mathbb{R}_l\times\mathbb{R}$ containing $(a,b)$, $\sigma(U)\not\subset V$. Now choose an open set
$$
V=[a+b,(a+b)+\tfrac{1}{2})\subset\mathbb{R}_l
$$
containing $(a,b)$. (You can choose any positive real number other than $\frac{1}{2}$.)
For any open set $U\subset\mathbb{R}_l\times\mathbb{R}$ containing $(a,b)$, we can find a basis element in $U$ satisfying
$$
(a,b) \in [a,a+\epsilon)\times(b-\delta,b+\delta)\subset U
$$
for some $\epsilon>0$ and $\delta>0$. Then
$$
\sigma\bigl([a,a+\epsilon)\times(b-\delta,b+\delta)\bigr)=\bigl((a+b)-\delta,(a+b)+\epsilon+\delta\bigr)\not\subset[a+b,(a+b)+\tfrac{1}{2})=V
$$
because the points less than $a+b$ are not contained in $V$. It follows that $\sigma(U)\not\subset V$. $\quad\square$

*Here is another proof showing that the addition $\sigma\colon\mathbb{R}_l\times\mathbb{R}\to\mathbb{R}_l$ is not continous. Assume that $\sigma$ is continuous. Then the image of the connected set $C=\{0\}\times(a,b)$ by $\sigma$ must be connected. But we have a contradiction that $\sigma(C)=(a,b)$ is not connected, because $(a,b)=(a,\frac{a+b}{2})\cup[\frac{a+b}{2},b)$ can be written as the disjoint union of two nonempty open sets. (In fact, $\mathbb{R}_l$ is totally disconnected.)
