Let $Y_1$ denote a 2-point discrete space, $Y_2$ a 2-point indiscrete space and $Y_3$ a 2-point space that is neither discrete nor indiscrete. Let $X$ denote a set with 3 points and describe topologies on $X$ such that $X$ has subspaces homeomorphic to two of the $Y_i$'s. Can there be a topology on $X$ such that $X$ has subspaces $Z_1$, $Z_2$, $Z_3$ such that $Z_i$ is homeomorphic to $Y_i$ for each $i$?
So I'm a little confused at how to conceptualize the $Y_2$ and $Y_3$ spaces. I'm also not sure how subspaces of X could be homeomorphic to two $Y_i$'s. Thanks for the help.
Best Answer
If $Y_1=\{a,b\}$ then the topology on $Y_1$ (the set of open sets) is $\bigl\{\emptyset,\{a\},\{b\},Y_1\bigr\}$. If $Y_2=\{c,d\}$ then the topology on $Y_2$ is $\bigl\{\emptyset,Y_2\bigr\}$. If $Y_3=\{e,f\}$ then the topology on $Y_3$ must be neither $\bigl\{\emptyset,\{e\},\{f\},Y_1\bigr\}$ nor $\bigl\{\emptyset,Y_3\bigr\}$; hence it is either $\bigl\{\emptyset,\{e\},Y_1\bigr\}$ or $\bigl\{\emptyset,\{f\},Y_1\bigr\}$ (and in fact there is no need to distinguish these two cases as swapping $e$ and $f$ is a homeomorphism between these two topologies); note that exactlya one of the points in $Y_3$ is closed.
Let $X=\{u,v,w\}$ be a three point space such that among the three two-point subspaces $Z_1=\{v,w\}, Z_2=\{u,w\},Z_3\{u,v\}$ there is one homeomorphic to each $Y_i$. Wlog. (i.e. because we may rename the elements of $X$) $Y_i\cong Z_i$. By definition of subspace topology,a set $U\subseteq Z_i$ is (relatively) open iff there is an open set $V\subseteq X$ with $U=V\cap Z_i$. From $Z_1\cong Y_1$, we see that one of the sets $\{v\}$ and $\{u,v\}$ is open. From $Z_2\cong Y_2$, we see that neither $\{u\}$ nor $\{u,v\}$ is open, hence $\{v\}$ is open. Also, one of $\{w\}$, $\{u,w\}$ is open (because of $Z_1$) but neither of $\{w\}$, $\{v,w\}$ is open, hence $\{u,w\}$ is open. But since $\{v\}$ and $\{u,w\}$ are open in $X$, their intersections $\{v\}$ and $\{u\}$ are relatively open in $Z_3$, thus making $Z_3$ discrete unlike $Y_3$.
Can we obtain a topology on $X$ if we drop the wish about $Z_3$? Using only info about $Z_1$ and $Z_2$, we have obtained the following info: $$\begin{align}\emptyset & \text{is always open}\\ \{u\}&\text{not open}\\ \{v\}&\text{open}\\ \{u,v\}&\text{not open}\\ \{w\}&\text{not open}\\ \{u,w\}&\text{open}\\ \{v,w\}&\text{not open}\\ \{u,v,w\}&X\text{ is always open}\\ \end{align} $$ That is, we already know that the topology on $X$ is precisely $\bigl\{\emptyset,\{v\},\{u,w\},X\bigr\}$.