Elementary set theory - are these sets empty?

Question

we are asked to answer if the following statements are true or false, and why:

1) The set ${\{\emptyset\}^{\mathbb N}}$ has exactly $1$ element.

2) The set ${{\emptyset}^{\mathbb N}}$ is empty.

3) The set ${\mathbb N}^{\emptyset}$ is empty

4) $(\{1,2,3\}^{\mathbb N})$-$\{1,2\}^{\mathbb N}$=$\{3\}^{\mathbb N}$

What I tried doing:

I tried using the fact that $|A^B|=|A|^{|B|}$, so for the first question we get $1^{\aleph0}$ = 1

For the second question, same thing, we get $0$, but for the third question, we have a problem...I think that set is empty, but when I look at $\aleph0 ^0$ it should be 1...

Answer 1

There is exactly one map $\mathbb N\to\{\emptyset\}$, given by $f(n)=\emptyset$.

There is no map $\mathbb N\to\emptyset$ as we cannot have e.g. $f(1)\in\emptyset$.

There is exactly one map $\emptyset\to\mathbb N$. If you give me an element of $\emptyset$, I am willing to name you a natural number.

For 4 just pick e.g. a surjection $\mathbb N\to\{1,2,3\}$.

Answer 2

Your reasoning about questions 1 and 2 is correct.

For question 3, note that for any set $X$, there exists a unique function $\emptyset \rightarrow X$, namely the empty function. This may seem counterintuitive at first, but keep thinking about it and it will start seeming more sensible & obvious. Anyway, the set of all functions $\emptyset \rightarrow X$ always has precisely one element, namely the empty function, and therefore $|X^\emptyset|=1.$ Let me reiterate: this becomes more intuitive the more you think about it.

I won't give you the answer to question 4, but here's a hint. You can think of $X^\mathbb{N}$ as the set of all sequences in $X$. Now consider the sequence $(1,2,3,1,2,3,1,\cdots) \in \{1,2,3\}^\mathbb{N}.$