Prob. 5 (c), Sec. 3, in Munkres' TOPOLOGY, 2nd ed: How to find this equivalence relation?

Question

Let $S$ be the following subset of the plane: $$S \colon= \{ \ x \times y \ | \ y = x+1, \ 0 < x < 2 \ \}.$$ Let $T$ be an equivalence relation on the real line such that $T$ is the intersection of all the equivalence relations on the real line that contain $S$.

Then how to describe the ordered pairs in $T$ completely?

My effort:

The following subset of the plane, which is an equivalence relation on the real line, contains $S$ as a proper subset: $$S^\prime \colon= \{ \ x \times y \ | \ y-x \in \mathbb{Z} \ \}.$$ So $$T \subset S^\prime.$$

For any $x \in (0,2)$, we have $x \times (x+1) \in T$, and so $(x+1) \times x \in T$ also, by virtue of symmetry. So, by transitivity, we can conclude that $x \times x \in T$ and $(x+1) \times (x+1) \in T$ for all $x \in (0,2)$.

What next?

Answer 1

Hint:

You are almost ready.

I preassume that in your question $x\times y$ is a notation for an ordered pair.

Also $(0,3)^2\cup\{x\times x\mid x\in\mathbb R\}$ is an equivalence relation that contains $S$.

So have a look at its intersection with $S'$.

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edit (theoretical part, practicizing my own notation for ordered pair)

For any $A\subseteq\mathbb{R}^{2}$ define $A^{op}=\left\{ \left(x,y\right)\mid\left(y,x\right)\in A\right\} $ and define $\triangle:=\left\{ \left(x,x\right)\mid x\in\mathbb{R}\right\} $.

A relation is reflexive if and only if it contains $\triangle$.

A relation is $Q$ symmetric if and only if $Q^{op}\subseteq Q$ (or equivalently $Q^{op}= Q$).

Note that $\left(A^{op}\right)^{op}=A$.

Then $T:=S\cup S^{op}$ is the smallest symmetric relation that contains $S$.

Now define $T^{n}$ by stating $T^{0}=\triangle$ and $T^{n+1}=T^{n}\circ T$.

Then the relation $E:=\cup_{n=0}^{\infty}T^{n}$ is transitive and is the smallest equivalence relation containing $S$.

Answer 2

Define an equivalence relation $\sim_S$ on $\mathbb{R}$ as follows: $$ \begin{align} &x \sim_S x &\:\:\text{ for $x \in \mathbb{R}$,} \\ &x \sim_S x+1 \sim_S x+2 &\:\:\text{ for $x \in (0, 1)$,} \\ &1 \sim_S 2 \text{.} \\ \end{align} $$ We show that $\sim_S = T$.

$\sim_S$ is well-defined, and is an equivalence relation which partitions $\mathbb{R}$ into equivalence classes of 1, 2 and 3 elements (for $x \notin (0,3)$, $x \in \{1,2\}$, and $x \in (0,1) \cup (1,2) \cup (2,3)$ respectively).

Furthermore, $S$ is contained in $\sim_S$: clearly, if $0 < x < 2$ then $x \sim_S \, x+1$, by considering cases $0 < x < 1$, $x = 1$, and $1 < x < 2$.

Thus by definition of $T$, we have $T \subseteq \sim_S$.

From the definition of $\sim_S$ it's evident that any equivalence relation $R\supseteq S$ has to contain $\sim_S$. So the intersection of all such equivalence relations contains $\sim_S$: that is, $\sim_S \subseteq T$.

Answer 3

For $x\leq 0$ or $x\geq 3$ let $E(x)=\{x\}$ . Let $[x]$ denote the largest integer not exceeding $x$ . For $x\in (0,1)\cup (1,2)\cup (2,3)$ let $E(x)=\{x-[x],x-[x]+1,x-[x]+2\} $ . Let $E(1)=E(2)=\{1,2\}$. Now let $T= \{E(x)\times E(x) : x\in R\}.$ Now let let $H$ be the set of equivalence relations $G$ on $R$ that satisfy $S\subset G . $ Then for every $x\in R$ we have $E(x)\subset \cap \{[x]_G :G \in H\}=[x]_T=E(x)$ because $T \in H$ and $[x]_T=E(x)$.

Answer 4

I am now reading Munkres and I had struggled with this problem for a long time. I know that it is an old post but I don't quite understand the answers written by Brian and Daniel (perhaps I am too dumb). So I want to give my own answer for anyone who will have trouble on this problem. My answer is quite similar to drhab's answer.

I will use the following notations:

• The "diagonal line" $D = \{(x,x)\,|\, x \in \Bbb{R}\}$.

• If $R$ is a relation on $\Bbb{R}$, let $E(R)$ be its equivalence closure, i.e. the smallest equivalence relation on $\Bbb{R}$ that contains $R$. Similarly, let $r(R)$, $s(R)$ and $t(R)$ be it reflexive closure, symmetric closure and transitive closure respectively. For simplicity, we shall write $r(s(S))$ as $rs(S)$ and $r(s(t(S)))$ as $rst(S)$.

Some more general comments about those closures:

• Given $S \subseteq \Bbb{R} \times \Bbb{R}$, $r(S)$, $s(S)$, $t(S)$ and $E(S)$ exist since $\Bbb{R} \times \Bbb{R}$ is a reflexive, symmetric and transitive relation that contains $S$ and we can take intersection as in Magidin's answer. So that $T = E(S)$.

• In particular, as noted in drhab's answer, $r(S) = S \cup D$ and $s(S) = \{(y,x)\, | \, (x,y) \in S\} \cup S$.

• In general, $st(S) \subsetneq ts(S)$. For $\subseteq$, $ts(S)$ is a transitive relation that contains $S$ so that $t(S) \subseteq ts(S)$. And $ts(S)$ is symmetric so that $st(S)\subseteq ts(S)$. But it can be the case that $st(S)$ fails to be transitive.

• Similarly, $rst(S), srt(S), str(S), rts(S), trs(S)$ and $tsr(S)$ are all subsets of $E(S)$. By showing $rts, trs, tsr$ satisfy reflexivity, symmetry and transitivity, $E(S) = rts(S)= trs(S)=tsr(S)$ but it can be the case that $rst(S) \subsetneq E(S)$, again it might fail to be transitive.

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From question 5(a), $S'$ is an equivalence relation on $\Bbb{R}$ containing $S.$

Let $R = D \cup (0,3)^2$. It is clear that $R$ is an equivalence relation on $\Bbb{R}$. Pick any $(x,y) \in S$. $x \in (0,2) \subseteq (0,3)$ and $y = x +1 \in (1,3) \subseteq (0,3)$. So $(x,y) \in (0,3)^2 \subseteq R$. So that $R$ is another equivalence relation that contains $S$.

We claim that $R \cap S' = E(S)$.

$R'\cap S \supseteq E(S)$: $R, S'$ are equivalence relations containing $S$, so that $R'\cap S$ is also an equivalence relation containing $S$ according to 5(b). Since $E(S)$ is the smallest equivalence relation that contains $S$, $E(S) \subseteq R'\cap S$.

$R\cap S'\subseteq E(S)$: Let $A = \{(x,x+2)\, |\, x \in (0,1) \}$ and $B = A \cup S$.

Claim 1: $S' \cap R \subseteq rs(B)$.

Pick any $(x,y) \in S' \cap R$. $y-x \in \Bbb{Z}$. But $x,y \in (0,3) \implies -3 < y-x < 3$.

Suppose $y - x = 0$. Then $x = y$, so that $(x,y) \in D\subseteq rs(B)$.

Suppose $y - x = 1$. But $y < 3 \implies x = y -1 < 2$. $x \in (0,2)$ and $y = x +1$. So that $(x, y) \in S\subseteq B \subseteq rs(B)$.

Suppose $y - x = 2$. Then $x \in (0,1)$ and $y = x+2$ so that $(x,y) \in A\subseteq B \subseteq rs(B)$.

For the cases $x - y = 1$ and $x-y = 2$, $(y, x) \in rs(B)$ according to the above two cases. But $rs(B)$ is symmetric so that $(x,y) \in rs(B)$.

Claim 2: $B \subseteq t(S) $

Pick any $(x,y) \in A$. Then $x \in (0,1)$. Then $x +1 \in (1,2) \subseteq (0,2)$ and $y = (x+1)+1$. We have $(x,x+1) \in S$ and $(x+1,y) \in S$. So that $(x,y) \in t(S)$ since $t(S)$ is transitive. So $A \subseteq t(S)$ and $S \subseteq t(S)$. Then $ A\cup S=B \subseteq t(S)$.

So that $rs(B) \subseteq rst(S)$. Combining this with claim 1, $S'\cap R\subseteq rs(B) \subseteq rst(S) \subseteq E(S)$.

This complete the proof. In this particular case, we have $rst(S) = E(S)=rs(B) = R\cap S'$. We can draw up $rs(B)$ or $R\cap S'$ to give Brian and Daniel's answer.