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I have a space $X$ with a subset, $A$, and then I let $Y = X \times \{0,1\} / R$ where $R$ is the equivalence relation defined by the partition $\{(a,0), (a,1)\}$ if $a \in A$ and $\{(x,i)\}$ for $i$ equal to $0$ or $1$. ($\{0,1\}$ has the discrete topology).

Now I am trying to show that if $Y$ is Hausdorff then $X$ is Hausdorff and $A$ is closed. I am using the fact that the graph of $R$, say $G \subset Y^2$, is closed. Now points $((x,i),(y,j)) \in G$ iff $x=y$ and ($x \in A$ or $i=j$). I also know that if we give $Y^2$ the product topology then $Y^2 -G $ is a union of sets $U\times V$ where $U$ and $V$ are open in $X \times \{0,1\}$.

But I am now really struggling to show that this gives me the required conclusion, I'm finding it hard to see what is actually going on with the graph of $R$. Thanks

Wooster
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  • what topology is on {0,1}? – Ryan Tran Mar 03 '14 at 22:25
  • Discrete topology, will add to question – Wooster Mar 03 '14 at 22:31
  • Sorry, i'm not sure i understand well the relation $R$: you say we have $(x,i)\sim(y,j)$ iff $i\in \lbrace 0,1 \rbrace$ and $x=y \in A$ or $(x,i)=(y,j)$? – Léo Mar 03 '14 at 22:34
  • So the relation is defined by the partition of pairs ${(x,0),(x,1)}$ if $x \in A$ and singletons ${x,i}$ if $x$ is not in $A$. So say if $X=[0,1]$ and $A = [1/4, 3/4]$ we would have $(1/2,0)$ is related to $(1/2, 1)$ but $(1/8,0)$ is only related to $(1/8, 0)$. Hope that makes sense! – Wooster Mar 03 '14 at 22:41
  • Duplicate of http://math.stackexchange.com/questions/346009/prove-that-y-is-hausdorff-iff-x-is-hausdorff-and-a-is-a-closed-subset-of?rq=1 ? – Amy Mar 03 '14 at 23:02

1 Answers1

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I would take a more pedestrian approach. Pick $x \in \overline{A}$. Let $V_i$ be a neighbourhood of $[(x,i)]$ in $Y$ (where $[(x,i)]$ denotes the equivalence class of $(x,i)$ modulo $R$). By the continuity of the projection, there are open neighbourhoods $U_i$ of $(x,i)$ in $X\times \{0,1\}$ that are mapped into $V_i$. Since $X\times\{i\}$ is open in $X\times\{0,1\}$, we may assume that $U_i$ is of the form $W_i \times\{i\}$ for an open neighbourhood $W_i$ of $x$. Let $W = W_0 \cap W_1$. Since $x \in \overline{A}$, there is an $a_W \in A\cap W$. But then $[(a,0)] = [(a,1)] \in \pi(W\times\{0\}) \cap \pi(W\times\{1\}) \subset V_0\cap V_1$, so $[(x,0)]$ and $[(x,1)]$ have no disjoint neighbourhoods, and since $Y$ is Hausdorff, that means $[(x,0)] = [(x,1)]$ or $x\in A$, hence $A$ is closed.

Now we show that the projection embeds $X\times \{0\}$ into $Y$, which then, since that space is homeomorphic to $X$, means $X$ is Hausdorff. Since the projection is continuous by definition of the quotient topology, and evidently injective on $X\times\{0\}$, it suffices to see that it is closed. Let $C \subset X$ be closed. Then $\pi^{-1}(\pi(C\times\{0\})) = C\times\{0\} \cup (C\cap A)\times\{1\}$. But $C\cap A$ is closed in $X$, so $\pi^{-1}(\pi(C\times\{0\}))$ is closed in $X\times\{0,1\}$, whence $\pi(C\times\{0\})$ is closed, and so $\pi\lvert_{X\times\{0\}}$ is closed, hence an embedding.

Daniel Fischer
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  • Thanks! This makes sense to me, strange, however, that the question actually suggests considering the graph of the relation as I have, so I assumed that way would probably be easier than jumping in with definitions. – Wooster Mar 03 '14 at 22:53
  • If we look at the graph of the relation and permute the factors a little, so view it as a subset of $(X\times X)\times {0,1}^2$, it becomes $$\Delta_X\times {(0,0)} \cup \Delta_X\times{(1,1)} \cup \Delta_A\times {(0,1)} \cup \Delta_A\times {(1,0)},$$ where $\Delta_M = {(m,m) : m\in M}$ is the diagonal of $M$. Since that is closed, you can deduce that $\Delta_A$ is closed in $X\times X$, and from that follows that $A = \delta^{-1}(\Delta_A)$ is closed in $X$, since $\delta\colon x \mapsto (x,x)$ is continuous. Maybe that's easier, I don't know. – Daniel Fischer Mar 03 '14 at 23:04