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Let $G$ be a group of homeomorphisms of a topological space $X$. The action of $G$ on $X$ is said to be discontinuous at a point $x \in X$ if

  1. $G_x :=$ the stabilizer of $x$, is finite.

  2. $x$ has an open neighbourhood $U$ such that $gU \cap U = \emptyset$ for all $g \notin G_x$

If $G$ is a topological group action on a topological space $X$ then the action is said to be continuous if the map $F : G \times X \rightarrow X$ given by $(g,x) \mapsto gx$ is continuous.

My question is - are these two notions the opposite of each other (considering the way they are named)? That is if $G$ is a topological group acting on $X$ then will the map $F$ not being continuous at some point $(x,g)\in G$ mean that the action is discontinuous at $x \in X$? I can't see why this should be true.

Can some one give me an example of a continuous group action that is discontinuous (if possible)?

Thank you.

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    $\begingroup$ look at $\mathbb{Z}$ acting on $\mathbb{R}$. $\endgroup$ – M.U. Sep 29 '15 at 8:55
  • $\begingroup$ It's "properly discontinuous", not "discontinuous". $\endgroup$ – YCor May 1 at 22:21
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"Discontinuous" is a weird choice of terminology for this, which is unfortunately standard. It is not the opposite of continuous in this context. In fact, one more-or-less never considers actions of topological groups that aren't continuous.

For an example, the action of $\mathbb{Z}$ on $\mathbb{R}$ by translation will do. (that is, for $n$ in $\mathbb{Z}$ and $x \in \mathbb{R}$, $n * x := x+n$). Stabilizers are trivial, and for a fixed $x$ in $\mathbb{R}$, to find $U$ as in your definition, you can take an open interval centered at $x$ of length less than $1$. Certainly this action, viewed as a map from $\mathbb{Z} \times \mathbb{R}$ to $\mathbb{R}$, is continuous (for instance because it is the restriction of the addition map $\mathbb{R} \times \mathbb{R} \to \mathbb{R}$.)

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