The definition of a $k$-manifold we are given is a set $M\subset\mathbb{R}^n$ such that the following equivalent conditions hold for each $x\in M$:
- There exists a mapping $f:\mathbb{R}^n\to\mathbb{R}^{n-k}$, continuously differentiable near $x_0$, such that $(Df)_{x_0}=A:\mathbb{R}^n\to\mathbb{R}^{n-k}$ is onto and $$ x\in M \iff f(x)=f(x_0)\text{ for all $x$ near $x_0$}$$
- There exists a local diffeomorphism $\varphi:\mathbb{R}^n\to\mathbb{R}^n$ near $x_0$, such that $$ x\in M \iff \varphi(x)\in \mathbb{R}^k\times\{0_{n-k}\}\text{ for all $x$ near $x_0$}$$
- There exists a permutation $(i_1,\dots,i_n)$ of ${1,\dots,n}$ and a mapping $g:\mathbb{R}^k\to\mathbb{R}^{n-k}$, continuously differentiable near $(x_{0,i_1},\dots,x_{0,i_n})$ such that $$ x\in M \iff g(x_{i_1},\dots, x_{i_k})=(x_{i_{k+1}},\dots,x_{i_n})\text{ for all $x$ near $x_0$}$$
The question is then simply
Which of the following subsets of $\mathbb{R}^2$ are 1-dimensional manifolds?
[...]
- $M_5=\mathbb{R}\times\{0,1\}$
- $M_6=\mathbb{R}\times\mathbb{Z}$
- $M_7=\mathbb{R}\times\{1,\frac{1}{2},\frac{1}{3},\dots\}$
Looking at $M_5$ I think it's not a manifold, as if I look at the permutation $(1,2)$ then as $(0,0),(0,1)\in M$, the function $g(x_1)=x_2$ needs to satisfy both $g(0)=0$ and $g(1)=1$.
Similarly the permutation $(2,1)$ fails as both $(0,0),(1,0)\in M$ which forces $g(x_1)=x_2$ to give both $g(0)=0$ and $g(0)=1$.
Is this reasoning correct? I assume not as then $M_6$ and $M_7$ would fail rather trivially, but what am I misunderstanding here?
Your mistake is that you ignored the word "near". If you take a small enough ball around $(0,0)$, this problem disappears.