Topologies on a 3 point space 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. 

Note that if $Y = \{ 1 , 2 \}$, then topologies on $Y$ corresponding to $Y_1$, $Y_2$, and $Y_3$, respectively, are


*

*$\{ \emptyset , \{ 1 \} , \{ 2 \} , Y \}$;

*$\{ \emptyset , Y \}$;

*$\{ \emptyset , \{ 1 \} , Y \}$.


Let us now consider a three element set $X = \{ a , b , c \}$.


*

*First suppose that we are given the following topology of $X$: $\{ \emptyset , \{ a \} , \{ b , c \} , X \}$.  Note that the subspace $\{ a , b \}$ of $X$ has as open sets the following:


*

*$\emptyset \cap \{ a , b \} = \emptyset$;

*$\{ a \} \cap \{ a , b \} = \{ a \}$;

*$\{ b , c \} \cap \{ a , b \} = \{ b \}$;

*$X \cap \{ a , b \} = \{ a , b \}$


And it follows that the $\{ a , b \}$ is a discrete subspace of $X$; i.e., it is homeomorphic to $Y_1$.

*Now consider the following topology of $X$: $\{ \emptyset , \{ b , c \} , X \}$.  In a similar manner to the above we find that the topology on the subspace $\{ a , b \}$ is $\{ \emptyset , \{ b \} , \{ a , b \} \}$, and so this subspace is homeomorphic to $Y_3$.

While you could go through all 29 topologies on $X$ (okay, 9 up to homeomorphism) to show that none has each of $Y_1 , Y_2, Y_3$ as subspaces, we can prove this by analysing those topologies on $X$ which have homeomorphic copies of both $Y_1$ and $Y_2$ as subspaces.
If $Z_1$, $Z_2$ denote subspaces of $X$ homeomorphic to $Y_1$, $Y_2$, respectively, note that $| Z_1 \cap Z_2 | = 1$.  So without loss of generality $Z_1 = \{ a , b \}$ and $Z_2 = \{ a , c \}$. Let $U,V,W$ denote the smallest open neighbourhoods of $a,b,c$, respectively, in $X$.


*

*As $U \cap Y_1 = \{ a \}$ and $U \cap Y_2 = \{ a , c \}$, we have $U = \{ a , c \}$.

*As $c \in U$ we must have $W \subseteq U = \{ a,c \}$, and as $W \cap Y_2 = \{ a , c \}$, it follows that $W = \{ a , c \}$.

*As $V \cap Y_1 = \{ b \}$, then  $V \subseteq \{ b , c \}$. If $c \in V$, then $V \cap W = \{ c \}$ is also an open neighbourhood of $c$, which is impossible since $W = \{ a , c \}$ is the smallest open neighbourhood of $c$. 


This means that the topology on $X$ is $\{ \varnothing , \{ a , c \} , \{ b \} , X \}$, and it is easy to show that $X$ has no subspace homeomorphic to $Y_3$.
