Let $X$ denotes the two-point set $\{0,1\}$ and write $X_j = \{0,1\}$ for every $j=0,1,2,\dots$. Let $Y = \prod\limits_{j=1}^{\infty} X_j$. Which of the following is/are true?

1) $Y$ is a countable set,

2) $\text{Card} \,Y = \text{Card} \,[0,1]$,

3) $\cup_{n=1}^{\infty}(\prod\limits_{j=1}^nX_j)$ is uncountable,

4) $Y$ is uncountable.

Claim: $\{0,1\}^{\Bbb N}$ is uncountable where $Y^X$ is the notation for the set of all functions from $X$ to $Y$.

Suppose this set is countable, then we can enumerate its elements and $\{0,1\}^{\Bbb N} = \{\overline x_1, \overline x_2, \overline x_3 \dots,\}$ where $\overline x_i$s are $0-1$ sequences. We will use Cantor's diagonalization argument to get a contradiction. We will construct a $0-1$ sequence $\overline y$ which is not listed above. This will prove that the above enumeration is not complete and hence the set $\{0,1\}^{\Bbb N}$ is uncountable. Define the ith term of the sequence $\overline y$ to be $\begin{cases}0 \text{ if the $i$th term of $\overline x_i$ is 1}\\ 1 \text{ otherwise }\end{cases}$. Now, this $0-1$ sequence $\overline y$ is different from the $\overline x_i$ in the $i$th position. So $\overline y \ne \overline x_i$ for any $i$ and hence $\overline y \notin \{0,1\}^{\Bbb N}$. Contradiction.

This shows that $Y$ is uncountable.

Click 1) $Y$ is a countable set,

2) $\text{Card} \,Y = \text{Card} \,[0,1]$,

3) $\cup_{n=1}^{\infty}(\prod\limits_{j=1}^nX_j)$ is uncountable,

4) $Y$ is uncountable.

**Solution**:**option 1**: (**False**)**option 4**: (**True**) Let $x$ be an element of $Y$, then $x = (x_1,x_2,x_3,\dots)$ an infinite tuple with entries $0$ and $1$. This can be identified with a $0-1$ sequence $\{x_n\}$ naturally.**This defines a bijection between the set $Y$ and the set of all $0-1$ sequences (which is denoted by $\{0,1\}^{\Bbb N}$)**.Claim: $\{0,1\}^{\Bbb N}$ is uncountable where $Y^X$ is the notation for the set of all functions from $X$ to $Y$.

Suppose this set is countable, then we can enumerate its elements and $\{0,1\}^{\Bbb N} = \{\overline x_1, \overline x_2, \overline x_3 \dots,\}$ where $\overline x_i$s are $0-1$ sequences. We will use Cantor's diagonalization argument to get a contradiction. We will construct a $0-1$ sequence $\overline y$ which is not listed above. This will prove that the above enumeration is not complete and hence the set $\{0,1\}^{\Bbb N}$ is uncountable. Define the ith term of the sequence $\overline y$ to be $\begin{cases}0 \text{ if the $i$th term of $\overline x_i$ is 1}\\ 1 \text{ otherwise }\end{cases}$. Now, this $0-1$ sequence $\overline y$ is different from the $\overline x_i$ in the $i$th position. So $\overline y \ne \overline x_i$ for any $i$ and hence $\overline y \notin \{0,1\}^{\Bbb N}$. Contradiction.

This shows that $Y$ is uncountable.

**option 2**: (**True**). Consider the binary representation of any element of $[0,1]$, this will be of the form $0.a_1a_2a_3\dots$ where $a_i$s are either $0$ or $1$ which can be identified with a $0-1$ sequence. This gives the required bijection.**option 3**: (**False**) Finite product of countable set is countable and a countable union of countable sets is countable.**This shows that the set given in option 3 is countable**. (Note that an infinite product of countable set is uncountable. Even infinite product of two-element set is uncountable which is proved in option 1)**here**for more problems.

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