Let $𝑌$ be a compact Hausdorff space. I want to show that there exists a homeomorphism between $𝑌$ and the subspace of Closed$(𝑌)$ which is a set of singleton sets.
Where I need to start with a function given by sing$:𝑌\rightarrow\text{Cts}(𝑌,\Sigma)$ such that $𝑦\mapsto\chi_{\{𝑦\}}^𝑐$ where $\chi$ is the characteristic function and sing is continuous.
I have proved that there exists a homeomorphism such that Cts$(𝑌,\Sigma)\cong$ Open$(𝑌)\cong$ Closed$(𝑌)$ such that I can define a function $𝑓$ which is the composite of those homeomorphisms.
I have defined a continuous inverse mapping $𝑔:\{\{𝑦\}|𝑦∈𝑌\}→𝑌$ with $\{𝑦\}\mapsto 𝑦$. $𝑔$ is obviously the inverse function but I want to show it is continuous.
Note that Closed($𝑌$) has the basis $\{\{𝑍\text{ closed in }𝑌|𝑍⊆𝑈\}\}$ where $𝑈$ is open in $𝑌$ and $𝑈^𝑐$ compact (which is automatically true as $𝑈^𝑐$ closed in compact $𝑌$).
Attempt: I have deduced that $𝑔^{−1}(𝑈)=\{\{𝑦\}∈ \text{singleton of Close}(𝑌)|𝑦∈𝑈\} = \{\{y\} \text{ closed in }Y|\{y\}\subseteq U\}$. And the subspace topology gives the fact that $\{\{𝑍\text{ closed in }𝑌|𝑍⊆𝑈\}\} \cap \{\{y\}|y\in Y\} = \{\{y\}|\{y\}\subseteq U\}$ hence $g$ is continuous. Am I correct? Thanks!
Best Answer
Discarding the weird notations and putting it in terms familiar to me:
You have a hyperspace of closed sets (which I usually call $H(Y)$) with a base all sets $U^+ = \{A \in H(Y)\mid A \subseteq U\}$ where we restrict to $U$ that are open and have compact complement in $Y$ (so half of the Fell topology) This is somewhat non-standard (I'm used to the Vietoris topology or the Fell topology), but seems to work for this question.
So $Y$ being Hausdorff and compact (just $T_1$ suffices), the map $i:Y \to H(Y); y \to \{y\}$ is well-defined.
It is continuous as $$i^{-1}[U^+] = \{y \in Y: \{y\} \subseteq U \} = U$$ for all basic open sets.
As $i$ is clearly injective and $i[U]=U^+ \cap i[Y]$ as well, $i$ is open as a map (any open $U \subseteq Y$ qualifies for usage as a base element as you already noted) from $Y$ onto $i[Y]$ and thus $Y \simeq i[Y]$ (and $i[Y]$ is the subspace if all singleton sets).
All other considerations are unneeded and just distract from the matter.