$\left\{0\right\}\times (0,1) \not \in \mathbb R^2$

Question

So I was taking my topology class and professor said it's trivial to see that $\left\{0\right\}\times (0,1)$ is not a member of the usual topology $ \mathbb R^2$, is there a way to prove this or is it really that trivial and I'm not seeing it?

Answer 1

The main idea is: a line segment in $\mathbb{R}^{2}$ can't contain an open ball.

To make this more rigorous, we'll expand on Randall's comments:

We claim that $\{0\} \times (0, 1)$ does not contain an open neighborhood containing the point $(0, \frac{1}{2})$.

To see this, note that any open neighborhood containing $(0, \frac{1}{2})$ contains some open ball $B((0, \frac{1}{2}), r)$ of radius $r > 0$ centered at $(0, \frac{1}{2})$. So, we just have to show that $\{0\} \times (0, 1)$ does not contain any open ball of positive radius centered at $(0, \frac{1}{2})$.

Note that an open ball of radius $r > 0$ centered at $(0, \frac{1}{2})$ contains the point $(\frac{r}{2}, \frac{1}{2}).$ This is because the distance between $(0, \frac{1}{2})$ and $(\frac{r}{2}, \frac{1}{2})$ is $\frac{r}{2}$, which is strictly less than $r.$ However, since $\frac{r}{2} > 0,$ it is clear that $(\frac{r}{2}, \frac{1}{2})$ is not in the set $\{0\} \times (0, 1).$

Since this argument works for any arbitrary $r > 0$, we see that $\{0\} \times (0, 1)$ does not contain any open ball of positive radius centered at $(0, \frac{1}{2})$, and so $\{0\} \times (0, 1)$ is not an open set.