Is this an open set: $B = \left \{ (x,y) \in \mathbb{R}^2| 0Suppose $$B = \left \{ (x,y) \in \mathbb{R}^2| 0<y<1 \right \} $$
I can see that it if I put a sequence in B such that $(x_n,y_n)=(n,1-\frac{1}{n})$, we can state that it's not a closed set because the limit of $(x_n,y_n) \not \in B$. So it's neither a compact nor a closed set. How can I finally show that it's open? 
 A: Take any $(x,y)$ in $B$ and prove that there exists some ball around $(x,y)$ that is still in $B$.
Hint: if $(x,y)\in B$, then $0<y<1$. Now, suppose $y<\frac12$. Is $\frac y2\in (0,1)$?
A: Hint: $B = \mathbb R \times (0, 1)$ which is open in box topology. Further,
$$B = \mathbb R \times (0, 1) = \bigcup_{a \in \mathbb R} \mathcal B((a, 1/2), 1/2),$$
where $\mathcal B((x, y), r)$ denotes an open ball on plane. Unions of open sets are open.
A: Draw the set $B$. Take a point $(x,y)\in B$. Then $0<y<1$. Draw it. You need to find a circle centered at $(x,y)$ contained in $B$. What would be its radius?
Hint: Consider separately two cases: $y<1/2$ and $y\ge 1/2$.
A: The function
$$f(x)=\frac1{\sqrt{2\pi}}\int_{-\infty}^xe^{-t^2/2}dt$$ is continuous. Then $B=f^{-1}(\Bbb R)$ is open.
A: U can think of it as the Intersection of two open sets, $\{(x,y)\in \mathbb R\ |\ y<1  \}\cap \{(x,y)\in \mathbb R\ |\ y >0 \}$, which both is easy two check they are open by a bijection with $\mathbb R$ or checking that they are Unions of balls, and we know that the intersection of finite open sets is open. 
