Consider the adjunction space $\Bbb RP^n \sqcup_p D^{n+1}$ which is obtained from the disjoint union $\Bbb R P^n \sqcup D^{n+1}$ by identifying $\xi \in S^n \subseteq D^{n+1}$ with $p(\xi),$ where $p : S^n \longrightarrow \Bbb RP^n$ is the quotient map. Show that the adjunction space $\Bbb R P^n \sqcup_p D^{n+1}$ is Hausdorff.
I know that both $\Bbb RP^n$ and $D^{n+1}$ are compact Hausdorff spaces. So the space $\Bbb R P^n \sqcup D^{n+1}$ is compact and Hausdorff and hence normal. But normal Hausdorff (in fact $T_1$ is enough) spaces are regular. I know that if $X$ is regular and $A$ is a closed subspace of $X$ then $X/A$ is Hausdorff. But what is $A$ here? Is there any nicer way of approaching this problem? Any suggestion regarding this will be appreciated.
Thanks for your time.
Given any two points $x,y$ in this adjunction space, there are three possibilities. If they are both in the embedding of the open disk, they admit separating neighborhoods, and if they are both in the embedding of $\mathbb{RP}^n$, they admit separating neighborhoods. This is because both of these spaces are Hausdorff. Finally, if $x$ is in the open disk and $y$ is in $\mathbb{RP}^n$, we can separate them as follows. Consider the set $U_{\delta}$, which is the open set whose preimage is the set $\{\lvert\lvert x\rvert\rvert>\delta\}$ in the disk. This will be our neighborhood of $y$. Take $\delta$ small enough that $\{\lvert\lvert x\rvert\rvert\geq\delta\}$ does not contain the preimage of $x$. Then we can choose a neighborhood of the preimage of $x$ disjoint from this set, and the image of this neighborhood gives us the desired neighborhood of $x$.
The method of collapsing a closed subspace does not apply here, because this quotient involves a more complicated equivalence relation.