I can't seem to find the mistake in this obviously false proof I've thought up while trying to understand topological groups. It's pretends to prove the discreteness of all topological groups.

Let $G$ be a topological group. Consider any set $A\subset G$. Then $A \ast G$ should equal $G$, given that for any $g \in G,a \in A$ we have $a^{-1}g \in G$ and $aa^{-1}g = g$.

Now given that $\ast : G \times G \to G$ is continuous and that $G$ is open that would imply that $A \times G$ is likewise open in the product topology and thus that $A$, too, is open.

But this last statement should not be true as far as I know.


$\ast^{-1}\left(G\right)=G\times G$ and not $A\times G$.

Note that while $\ast\left(A\times G\right)=G$, $\ast\left(G\times G\right)=G$ as well and the inverse image is defined as ALL $(g_1,g_1)\in G\times G$ such that $\ast(g_1,g_2)\in G$.


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