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.

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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 $\bigl\{\emptyset,\{v\},\{u,w\},X\bigr\}$.


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

  1. $\{ \emptyset , \{ 1 \} , \{ 2 \} , Y \}$;
  2. $\{ \emptyset , Y \}$;
  3. $\{ \emptyset , \{ 1 \} , Y \}$.

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 $Y_3$.


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