TITLE: Collection of Equivalence relations
QUESTION [0 upvotes]: 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 $(x,z) \in R$.

REPLY [2 votes]: 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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