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I believe I have these all correct, but if I made an error could you lend a hand and possibly explain why I was mistaken? Thanks in advance.

Consider the following sets:

$X_{1}=\emptyset$

$X_{2}=\mathbb{N}$

$X_{3}=\mathbb{R}$

$X_{4}=\{\mathbb{N},\mathbb{R}\}$

$X_{5}=\{R|R$ is a relation on $\{1,2,3,4,5\}\}$

$X_{6}=\{R|R$ is a relation on $\mathbb{N}\}$

$X_{7}=\{R|R$ is a relation on $\mathbb{R}\}$

$X_{8}=\mathbb{R}\times\mathbb{R}$

$X_{9}=\{0,1\}^\mathbb{N}$

$X_{10}=\mathbb{N}^{\{0,1\}}$

$X_{11}=\mathbb{Q}\times\mathbb{R}$

$X_{12}=\mathcal{P}(\{\emptyset,\mathbb{N},\mathbb{R}\})$

$X_{13}=\{0,1\}^\mathbb{R}$

a) Which sets above are finite?

$X_{1}$,$X_{4},X_{12}$,$X_{5}$

b) For the sets which are finite, order then by ascending cardinality.

$X_{1}$,$X_{4},X_{12}$,$X_{5}$

c) Which sets above are countable?

$X_{1}$,$X_{2}$,$X_{4}$,$X_{5}$,$X_{6}$,$X_{10},X_{12}$

d) Which sets above are countably infinite?

$X_{2}$,$X_{6}$,$X_{10}$

e) Which sets above are uncountable?

$X_{3}$,$X_{7}$,$X_{8}$,$X_{9}$,$X_{11}$,$X_{13}$

f) Which sets have cardinality greater than c = |R|?

$X_{7}$,$X_{13}$

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    $\begingroup$ This is gonna take very long to answer, but just glancing, you are missing $X_4$ for finiteness. The question asks for the particular set, not the objects. $\endgroup$ – IAmNoOne Jun 21 '16 at 1:26
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    $\begingroup$ $X_{12}$ is finite too, for similar reasons. $\endgroup$ – lulu Jun 21 '16 at 1:29
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    $\begingroup$ $X_{6}$ isn't finite. Let $f : \mathbb{N} \to \mathbb{N}$ be a bijection, and let $R_f$ be defined on $\mathbb{N}$ be defined by $(a R_f b) \iff \left( f^{-1}(a) \leq f^{-1}(b) \right)$. You should be able to construct infinitely many such bijections without much difficulty. $\endgroup$ – AJY Jun 21 '16 at 1:47
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    $\begingroup$ Part $f$ is wrong. $X_8$ has the same cardinality as $\mathbb R$. $\endgroup$ – Matt Samuel Jun 21 '16 at 2:08
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    $\begingroup$ @shai Since your comment got an upvote I think it's important to point out that what you say is not true. $\mathbb N^{\{0,1\}}$ is the set of all functions from a two element set into the natural numbers and hence is countable. $\endgroup$ – Matt Samuel Jun 21 '16 at 2:34

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