Open set question So the set $\{(x,y)\in\Bbb R^2  \mid y=0,\; 0 < x < 1\}$ is not open in $\Bbb R^2$ correct? Since there is no neighborhood around $y=0$ that is in the set. Would $\{(x,y)\in\Bbb R^2  \mid y=0,\; 0 \le x \le 1\}$ not be open for similar reasons? For the sets $X=\{(x,y)\in\Bbb R^2 \mid x^2+y^2\le 1\}$ and $Y=B(0,1)\cup\{(1,0)\}$ is $Y$ open in $X$? I'm a little confused about demonstrating that a set is open and looking for someone to clear things up, thanks!
 A: I assume that you mean that an "open sets $S$ in $\mathbb{R}^2$" are those sets with the property that for any $p \in S$, there is some radius $r$ such that an open ball of radius $r$ around the point $p$, denoted by $B_r(p)$, is contained in $S$.
If this is the case, then your questions about line segments not being open are correct. In particular, no ball sits in a line. But you ask for a neighborhood around $y=0$, which is a line and not a point. Perhaps you meant around the point $(1/2, 0)$, which is on your line segment.
If $B(0,1)$ is an open ball around $0$ with radius $1$, then $B(0,1)\cup \{(1,0)\}$ is not open in $X$, because a ball around the point $(1,0)$ would contain some other part of the circumference around the ball $B(0,1)$ that is contained in $X$ but that is not contained in $B(0,1)$. (I assume that for subspaces $U$ of $\mathbb{R}^2$, we call sets open in $U$ if they are an open set in $\mathbb{R}^2$ that is restricted to $U$, often referred to as the subspace topology).
A: Define $S = \{(x,y)\in \mathbb{R} | y=0, 0 < x <1$ }.  Then $(\frac{1}{2}, 0)\in S$ and given $\epsilon> 0$ the point $(\frac{1}{2}, \frac{\epsilon}{2})\in B_{\epsilon}(\frac{1}{2},0)$ but $(\frac{1}{2},\frac{\epsilon}{2})\not\in S.$ Hence there is no open neighborhood around $(\frac{1}{2},0)$ that is contained in $S$ and thus $S$ is not open.
The same exact proof will work for your second set and the same reasoning will work for the third set
