Left Semigroups and Compact sets I have fallen down the rabbit hole of reading old papers. I (believe that I) have read the following claim: 
Let $(S,\cdot)$ be a semigroup and assume that $S$ is also a topological space such that for each $b \in S$, the map $- \cdot b: S \to S$ is a continuous map. Let $A,B \subset S$ such that $A,B$ are compact. Then, $A \cdot B$ is compact. 
I might just be getting old and rusty, but I can't seem to show this. Do I need to assume that $\cdot$ is jointly continuous? In particular, that the maps $a \cdot -$ are continuous? 
Edit: I've added the [model-theory] tag since I assume someone here has had thoughts on this before in relation to the Ellis Semigroup. 
 A: This isn't true under the hypotheses you gave. For example, let $S$ be the real numbers with their usual topology, and define $a\cdot b = 0$ if $b$ is an integer and $a\cdot b = ab$ (usual multiplication) otherwise. Then $(-\cdot b)$ is continuous for any fixed $b$, but $[0,1]\cdot[0,1]=[0,1)$. 

Edit: As J.-E. Pin points out in the comments, the example I gave above is not actually a semigroup. Here's an example that actually works:
Let $S = \mathbb{R}\cup \{\infty\}$ be the one-point compactification of $\mathbb{R}$ (the circle), with a semigroup operation defined by: $$a\cdot b = \begin{cases} 0& \text{if }b = \infty\text{ or }b = 0\\ ab & \text{if }a\in \mathbb{R} \text{ and }b\in \mathbb{R}^\times\\ \infty &\text{if }a = \infty\text{ and }b\in \mathbb{R}^\times. \end{cases}$$
For any $b$, $(-\cdot b) \colon S\to S$ is continuous. This map is constant if $b = \infty$ or $b = 0$, and otherwise it is the "multiply by $b$" map $\mathbb{R}\to \mathbb{R}$ extended continuously to a map $S\to S$. And $\cdot$ is associative: it is easy to check that $$(a\cdot b)\cdot c = a\cdot (b\cdot c) = \begin{cases} abc & \text{if }a,b,c\in \mathbb{R}^\times\\
\infty & \text{if }a = \infty \text{ and }b,c\in \mathbb{R}^\times\\
0 & \text{otherwise}.\end{cases}$$
The sets $\{1\}$ and $S$ are compact, but $\{1\}\cdot S = \mathbb{R}$, which is not compact.
