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I'm lazy about retyping the question so I put a screenshot of it here: Munkres, Topology, Chapter 1, section 3, problem 5(b).

Let $\{R_i\}_{i \in I}$ be a collection of equivalence relations with index set $I$ on a set $A$. Let $R=\cap_{i\in I} \{R_i\}$. I want to show that this is again an equivalence class.

To show $R$ is an equivalence class I need to show that it is reflexive, symmetric, and transitive.

Reflexive: I attempted this one but got some circular reasoning.

Symmetric: Let $(x,y) \in R$. Then $(x,y) \in R_i$ for all $i \in I$. Since each $R_i$ is an equivalence relation on $A$, $(y,x) \in R_i$ for all $i$. So $(y,x) \in R$.

Transitive: Let $(x,y), (y,z) \in R$. Then $(x,y), (y,z) \in R_i$ for all $i \in I$. Since each $R_i$ is an equivalence relation on $A$, $(x,z)$ is in each $R_i$. So $(x,z) \in R$.

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up vote 2 down vote accepted

But it's exactly the same reasoning as with symmetry and transitivity!

$$\forall\,a\in A\,\,\wedge\,\,\forall\,i\in I\,\,,\,\,(a,a)\in R_i\Longrightarrow (a,a)\in\bigcap_{i\in I}R_i$$

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This is so elegant and simple. I just keep reading it over and over again. Is the rest of my proof ok? – emka Jan 24 '13 at 4:28
I think you mean that for every $a \in A$, we have $\forall i \in I\,(a,a)\in R_i$ and therefore $(a,a) \in \bigcap_{i\in I} R_i$. – Trevor Wilson Jan 24 '13 at 5:30
Yes, it is..... – DonAntonio Jan 24 '13 at 5:32
@TrevorWilson , isn't that what I wrote? – DonAntonio Jan 24 '13 at 5:34
What you wrote isn't grammatical because the part of the left of "$\wedge$" is not a formula. But also on the right of "$\wedge$" I think you may want parentheses around "$\forall i \in I\, (a,a)\in R_i$". – Trevor Wilson Jan 24 '13 at 5:39

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