Why is the set $(0,1)\times \{0\}$ not open with respect to $\mathbb{R}^2$?

Question

So $\lbrack 0, 1 \rbrack\times \{0\}$ is closed with respect to $\mathbb{R}^2$. However, $(0,1)\times \{0\}$ is neither open nor closed with respect to $\mathbb{R}^2$. Why is this? I get why it is not closed, since the points $(0,0)$ and $(1,0)$ are limit points of the set but not contained in the set. However, why is it not open?

Answer 1

In $\mathbb R^2$ an open ball is... well an open disk and it surrounds its center not merely in the $x $ direction but in the $y $ direction as well.

As $A=(0,1)\times\{0\} $ has no "give" in the $y $ direction-- every point in $A $ has a precise $y$ value of exactly $0$--and we can't "fit" a neighborhood around it as a neighborhood would extend beyond $0$ in the $y $ direction.

Formally: for any $(x,0)\in A $ and any possible $\epsilon >0$ then the point $(x,\frac \epsilon 2) $ will be in $B_\epsilon ((x,0))$, the neighborhood of $(x,0) $ with radius $\epsilon$, but $(x,\frac \epsilon 2) $ isn't in $A $. So $B_\epsilon ((x,0))\not \subset A$, and so $(x,0) $ is not an interior point.

Answer 2

A basic open neighborhood in $\mathbb{R}^2$ around, say, $(0.5,0)$ will necessarily leave the interval since it is an open ball and NOT an open interval.

Answer 3

For it to be open, every point has to be an interior point. But none of its points are interior points, because you can't draw an open ball around any of its points that is contained in itself.

Answer 4

Take any point $(x, 0)$, where $0 < x < 1$, and fix $\varepsilon > 0$. Then the point $(x, \varepsilon / 2)$ iss less than $\varepsilon$ distance away from $x$, but does not lie in $(0, 1) \times \lbrace 0 \rbrace$. Therefore, no matter how small we make $\varepsilon > 0$, there is never a ball $B((x, 0); \varepsilon)$ contained in $(0, 1) \times \lbrace 0 \rbrace$.

In fact, we've shown that the set has empty interior (and it's non-empty) so it cannot be open.

Answer 5

Take any point, like $x=(\frac12,0)$. There's no neighborhood $N\subset \mathbb R^2$ with $(\frac12,0)\subset N\subset (0,1)×\{0\}$.

For, there would have to be a ball $B(x,\delta)\subset N$. But such a ball must contain a point $(y,z)$ with $z\not=0$.