Relatively Open Subset, Proof

Question

The exercise is:

Let $ A \equiv [0, 1). $ Prove that $ B \subset A $ is relatively open in A only if it is open in $ \mathbb{R} $ or it contains $0$.

Context: We're talking metric spaces with a distance $ d$.

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I wanted to ask if my proof is watertight:

Let $ A \equiv [0, 1) $ and $ B \subset A.$

WTS if $B$ is relatively open in $A$ then $0 \in B$ or $B$ open in $\mathbb{R}$.

Assume $ B \subset A $ is relatively open in $A$.

Then $ \forall x \in B , \ \exists \varepsilon>0$ such that $\{ y \in A | d(x,y)<\varepsilon \} \subset B.$

Assume $0 \not\in B$, so $B \subset \text{int}(A) $ where $\text{int}(A) \equiv \{ z \in A |\exists \varepsilon > 0, B_{\varepsilon}(z) \subset A \}. $

Since $\forall x \in B, \exists \varepsilon>0$ such that $\{ y \in A | d(x,y)< \varepsilon \} \subset B\subset \text{int}(A)$ and $\text{int}(A)$ is an open set in $\mathbb{R}$, it is the case that $\forall x \in B, \ \exists \varepsilon > 0$ such that $\{ y \in \mathbb{R} | d(x,y) < \varepsilon \} \subset B.$

Therefore, $B$ is open in $\mathbb{R}$.

Thanks in advance.

Answer 1

Since B is relatively open, there is an open set U, of R with B = A $\cap$ U.
There are two cases: 0 in B and 0 not in B.

In the second case, B is a subset of (0,1).
Thus B = (0,1) $\cap$ U, is the intersection of two open subsets of R, hence an open set of R.