Are $\{\frac{1}{n}|n∈\mathbb{N}\}\cup\{0\}$ and $\mathbb{Z}$ homeomorphic? Would anyone be able to explain why not?
I feel like you can show it using sequences, but I'm not sure how.
Usual metric of $\mathbb{R}$.
 A: $0$ is a limit point in the first set.  But $\mathbb Z $ has none.
A: The intuition is in the comments:

One has a limit point, the other doesn't

So let's run with this and write a proof by contradiction: We'll show that any bijection $$h: \bigg\{\frac{1}{n}\ \bigg|\ n=1,2,\ldots\bigg\}\cup\{0\} \to \mathbb{Z}$$ cannot be continuous. We will apply the intuition by focusing on the limit point $0$ and its image under $h$.
For convenience denote $X = \bigg\{\frac{1}{n}\ \bigg|\ n=1,2,\ldots\bigg\}$.
Let $p = h(0)$. The set $\{p\}$ is open in $\mathbb{Z}$. But its preimage $h^{-1}(\{p\}) = \{0\}$ is not open in $X$ because every open set of $X$ that contains $0$ also contains at least one other point.
Since a homeomorphism is, in particular, a continuous bijection, there is no homeomorphism between $X$ and $\mathbb{Z}$.

Bonus material
An accumulation point of a topological space $X$ is a point $p\in X$ such that every open set $U$ containing $p$ also contains at least one other point of $X$, that is, for every open $U\ni p$ there exists $q\in X$ with $q\neq p$ and $q\in U$.
Exercise 1: Let $X,Y$ be topological spaces and let $f:X\to Y$ be continuous and injective. Show that if $p\in X$ is an accumulation point of $X$, then $f(p)$ is an accumulation point of $Y$.
Exercise 2: Apply Exercise 1 to your question :)
