Let $A = [0,1]$. Determine whether these sets are open in $A$.

Question

Let $A = [0,1] \subset \mathbb R$. Determine which of the following subsets of $A$ are open in $A$: $\left({1\over2}, 1\right], \left({1\over2},1\right), \left[{1\over2},1\right)$.

I'm having a little bit of a difficult time wrapping my head around sets being open with respect to another metric space. By definition, I know that

Let $(M, \rho)$ be a metric space with $A\subset M$. Then $G_A\subseteq A$ is an open subset of $(A,\rho)$ if and only if $G_A = A\cap G_M$, where $G_M$ is an open subset of $M$.

For $\left({1\over2},1\right]$, we can easily let $G_{\mathbb R} = \left({1\over2},2\right)$, which is definitely open in $\mathbb R$. So, $$G_A = \left({1\over2},2\right)\cap[0,1] = G_{\mathbb R}\cap A = \left({1\over2},1\right].$$ Hence $G_A$ is open in $A$.

The same argument I believe can be applied to $\left({1\over2},1\right)$. Just take $G_{\mathbb R} = G_A = \left({1\over2},1\right)$ and we have that $G_A$ is open in $A$ since open intervals are open in $\mathbb R$.

I'm having a little trouble with the third interval. The interval $\left[{1\over2},1\right)$ is not open in $A$. I'm not sure how to deduce this. I would need some $G_{\mathbb R}$ open in $\mathbb R$ such that $G_A = A \cap G_{\mathbb R} = [0,1]\cap G_{\mathbb R} = \left[{1\over2},1\right).$ But I can't imagine any set $G_{\mathbb R}$ that would be open in $\mathbb R$ fitting this criteria. Is there some systematic way of finding such a set if it exists (Likely it doesn't).

An additional question I am asked to determine is

Are there are any subsets of $\mathbb R$ that are open in $\mathbb R^2$?

The answer to this question I am inclined to say that there are no such subsets. Subsets of $\mathbb R$ are $1D$, and hence you can't fit a $2D$ ball inside a $1D$ line. But this doesn't really use the definition above. Can anyone provide some guidance as to how to solve these problems?

Answer 1

$B=[\frac{1}{2},1)$ is not open in $A$, since $\frac{1}{2}$ is a boundary point of $B$ which belongs to $B,$ so any neighborhood of $\frac{1}{2}$ intersects $A \setminus B$ and thus is not entirely contained in $B.$

If $U \subset \mathbb{R}$ is not empty, $U$ is not open in $\mathbb{R}^2$ since any (non empty) open in $\mathbb{R}^2$ must contain an open ball and clearly a non empty subset of $\mathbb{R}$ can't contain a ball of $\mathbb{R}^2.$

Answer 2

Your answer to the first two sets are correct. Now for some tips on the third one.

To show that $\left[\frac{1}{2},1\right)$ is open in $[0,1]$, you need to find some open subset $U$ of $\mathbb R$ such that $\left[\frac{1}{2},1\right)=U\cap[0,1]$. To show it is not open, you need to show that for any open subset $U$ of $\mathbb R$, $\left[\frac{1}{2},1\right)\neq U\cap[0,1]$.

HINT Recall how the topology of $\mathbb R$ is defined, and focus on the point $\frac{1}{2}$.