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What is the cardinality of all equivalence relations in $\mathbb{R}$, such that their equivalence classes contain only finite,odd number of elements?

My attempt: as relations are subset of $\mathbb{R} \times \mathbb{R}$, so the cardinality is smaller than $2^{\mathbb{R}}$. So we need to prove that the cardinality of those relations is at least $2^{\mathbb{R}}$. One way could be to find an injective function from power set of R to those relations, and thus prove the statement.

But I have no idea how to construct such a function.

Afterall how should I approach such kind of questions? Are there any heuristics for solving such problems (constructing such functions)?

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To get $2^{\mathfrak c}$ you need to find continuum many binary choices to make. Assuming AC, you can divide $\Bbb R$ into continuum many disjoint three point sets. Now each set can consist of one equivalence class or three.

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    I don't understand what "disjoint three point sets" means. Could you show an example of that kind of partition? – 2017-01-15
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    No, I can't. There are continuum many sets, too many to list. Each set contains three real numbers and each real number is in exactly one of the sets. – 2017-01-15
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    OK I see. So, a given equivalence relation you identify with a partition of R, thus showing that the relations are continuum. Is that correct? – 2017-01-15
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    No, I identify each relation with a subset of $R$, so there are $2^\{mathfrak c$ of them. There is a bijection between my sets and $\Bbb R$. For any subset of $\Bbb R$, if $x$ is in it you make the corresponding one of my sets into three equivalence classes of one number each. If $x$ is not in it you make the set into one class of three numbers. This gives $\mathfrak c$ many binary choices, so $2^{\mathfrak c}$ relations. – 2017-01-15
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    What about the second part of the question? How to approach problems like this one? – 2017-01-15
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    This shows one approach. Think about how many choices you need, then find a way to get that many choices. If you find a small set, here three reals, that gives you a binary choice, you get lots of choices. – 2017-01-15