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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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Since you are new, I want to give some advice about the site: To get the best possible answers, you should explain what your thoughts on the problem are so far. That way, people won't tell you things you already know, and they can write answers at an appropriate level; also, people tend to be more willing to help you if you show that you've tried the problem yourself. If this is homework, please add the [homework] tag; people will still help, so don't worry. – Zev Chonoles Feb 19 '13 at 6:10
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Thanks for the suggestions and help – user62931 Feb 19 '13 at 6:15
@Zev: I'm all for not writing questions in imperatives, but simply asking a question without explicitly requesting help isn't "demanding an answer". I think it's quite acceptable, though perhaps not the most interactive of styles, to just pose a question, and I don't want all questions extended by a formal request for help that adds nothing to the content. Also I don't remember ever seeing anyone criticize such a no-frills question as rude. – joriki Feb 19 '13 at 7:35
@joriki: I suppose that's a fair distinction; and in general I think I should make a more nuanced array of standard posts for newcomers. I'll remove that part from my comment above. – Zev Chonoles Feb 19 '13 at 7:57

2 Answers 2

up vote 2 down vote accepted

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

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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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