Showing that the set $\mathbb{R^2} \setminus \{(n,0) \mid n \in \mathbb{N} \}$ is an open subset of the plane. 
For every $n \in \mathbb{N}$ denote $x_n=(n,0) \in \mathbb{R^2}.$ Show that the set $\mathbb{R^2} \setminus \{x_n \mid n \in \mathbb{N} \}$ is an open subset of the plane.

The set $\mathbb{R^2} \setminus \{x_n \mid n \in \mathbb{N} \}$ is the plane excluding the positive $x-$axis? It seems that I cannot use the definition of a ball here to conclude that the set would be open. Other definition that I know states that the union of open sets is open also, but it doesn't seem applicable here also. What other definitions might I use here?
 A: First of all, the most direct way to proceed in these situations is exactly what hamam_Abdallah proposed in his answer. Recall that closed sets are by definition complements of open sets, so if you manage to do that, it will immediately follow that $\mathbb{R}^2 \setminus (x_n)_{n \in \mathbb{N}}$ is open. Indeed, let $(a_k)_{k \in \mathbb{N}} \subseteq \{x_n: n \in \mathbb{N}\}$ be a converging sequence and denote its limit by $\lambda.$ Since convergence in the topology of $\mathbb{R}^2$ is the same as convergence on components, we deduce that $\lambda = (\mu, 0)$ for some $\mu \in \mathbb{R}.$ You should be familiar with the fact that a converging sequence with elements in $\mathbb{Z}$ is constant from a certain point on, but for the sake of completeness, I will provide a proof of this fact here. Note that this would complete our endeavour since it would follow that $\mu = n$ for some $n \in \mathbb{N},$ which would enable us to conclude that $(x_n)_{n \in \mathbb{N}}$ is closed (or, equivalently, that its complement is open). Thus, let $(y_n)_{n \in \mathbb{N}} \subseteq \mathbb{Z}$ be a converging sequence and let $\lambda$ be its limit. By the definition of limit, this means that there is $n_0 \in \mathbb{N}$ such that $y_n \in (\lambda - \frac{1}{2}, \lambda + \frac{1}{2})$ for any $n \geq n_0.$ But any such open interval of length $1$ contains at most one integer, so this interval contains exactly one integer $k$ and that integer must in fact be equal to $\lambda$ seeing as $y_n = k$ for all $n \geq 0.$ Thus, we proved our claim.
However, there is another more direct way to approach this problem (i.e. directly from  the definition of what being an open set means). Let $x \in \mathbb{R}^2 \setminus (x_n)_{n \in \mathbb{N}}$ and write $x = (x_1, x_2).$ Let $f := \lfloor x_1 \rfloor$ and $r := \frac{1}{2} \min \{d(x, (f, 0)), d(x, (f+1, 0))\},$ where $d$ stands for the standard euclidian distance. Then we infer that the ball $B(x, r)$ is included in $\mathbb{R}^2 \setminus (x_n)_{n \in \mathbb{N}}.$ Thus, the set $\mathbb{R}^2 \setminus (x_n)_{n \in \mathbb{N}}$ is open.
A: Consider the function $\mathbb R^2 \to \mathbb R^3$ given by $f(x,y)=(\sin(x \pi),x,y)$. Then $f$ is continuous.
Let $C=\{0\} \times [0,\infty) \times \{0\}$. Then $C$ is closed in $\mathbb R^3$.
Therefore, $f^{-1}(C)=\mathbb N \times \{0\}$ is closed in $\mathbb R^2$ and so its complement is open.
(I assume that $0 \in \mathbb N$. Otherwise, take $[1,\infty)$ instead of $[0,\infty)$.)
A: As $\Bbb N$ and $\{0\}$ are closed in $\Bbb R$, $\Bbb N \times \{0\}$ is closed in $\Bbb R^2$ and so your set, which is the complement of that set, is open.
