If a Hausdorff space $\ X\ $ admits a dense subset $ A \hookrightarrow X\ $ such that


$$|X|^{|A|}\ =\ |X|$$


then indeed $$|X| \leq |\text{End}\_{\text{Top}}(X)| \leq |X|^{|A|}\ = \ |X|.$$


It is the case of $\mathbb{Q} \hookrightarrow \mathbb{R}$. Thus, if there is a small enough dense subspace, there are not so many endomorphisms. 



> 
> Is the converse true?
> 
> 
> 


Namely, suppose $|\text{End}\_{\text{Top}}(X)| = \ |X|.$ Is it true that there exists a dense subset $A$ such that $|X|^{|A|}\ =\ |X|$?




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This question generated some attention on [MSE](https://math.stackexchange.com/questions/2533941/some-t2-spaces-must-have-a-small-dense), but it did not receive any answer, thus I am reposting it here.



Some T2 spaces must have a small dense?
**Is the converse true?**


No.


The paper *[Constructions and Applications of Rigid Spaces, I](http://dx.doi.org/10.1016/0001-8708(78)90006-3)*, Advances in Mathematics 29 (1978), 89-130, by Kannan and Rajagopalan, describes a countaby infinite Hausdorff space $X$ such that the only continuous maps $f\colon X\to X$ are the constant maps and the identity map. Such a space satisfies 
$|\text{End}\_{\text{Top}}(X)| = \omega = \ |X|.$ 


If $A\subseteq X$ is a subset such that $|X|^{|A|}=|X|=\omega$, then $A$ is forced to be finite, and therefore $A$ cannot be dense.


