> 
> $(X',\tau')$ is a compactification of $(X,\tau)$ if
> 
> 
> 1. $(X',\tau')$ is compact
> 2. $(X,\tau)$ is a topological subspace of $(X',\tau')$
> 3. $X$ is dense in $X'$.
> 
> 
> 

I was wondering why it is necessary to have the last condition. In my notes, I have written down that the third point is there to prevent 'trivial' compactifications such as $\bar{X}^{(X')}$, i.e. the closure of $X$ in $X'$. However, I don't see how this last one is compact.

So, my main question would be: why are we requiring that $X$ is dense in $X'$?