Consider the set \(Y = [-1, 1]\) as a subspace of \(\mathbb{R}\). Which of the following sets are open in \(Y\)? Which are open in \(\mathbb{R}\)?
\[ A = \ts{x \mid \tfrac{1}{2} < \abs{x} < 1}, \]
\[ B = \ts{x \mid \tfrac{1}{2} < \abs{x} \leq 1}, \]
\[ C = \ts{x \mid \tfrac{1}{2} \leq \abs{x} < 1}, \]
\[ D = \ts{x \mid \tfrac{1}{2} \leq \abs{x} \leq 1}, \]
\[ E = \ts{x \mid 0 < \abs{x} < 1 \text{ and } 1/x \notin \mathbb{Z}_+}. \]
Solution
Write the sets explicitly: \[ A=(-1,-\tfrac12)\cup(\tfrac12,1), \] \[ B=[-1,-\tfrac12)\cup(\tfrac12,1], \] \[ C=(-1,-\tfrac12]\cup[\tfrac12,1), \] and \[ D=[-1,-\tfrac12]\cup[\tfrac12,1]. \] Thus \(A\) is open both in \(\mathbb R\) and in \(Y=[-1,1]\). The set \(B\) is open in \(Y\), since \[ [-1,-\tfrac12)=Y\cap(-2,-\tfrac12),\qquad (\tfrac12,1]=Y\cap(\tfrac12,2), \] but it is not open in \(\mathbb R\): it contains \(1\), and every open interval about \(1\) contains points greater than \(1\). The sets \(C\) and \(D\) are not open even in \(Y\), since the included points \(\pm\tfrac12\) have no sufficiently small relative neighborhood contained in the set; consequently they are not open in \(\mathbb R\) either.
Finally \[ E=(-1,0)\cup\left((0,1)-\{1/n:n\ge2\}\right). \] The only accumulation point of \(\{1/n:n\ge2\}\) in \(\mathbb R\) is \(0\), which is not in \(E\). Hence every point of \(E\) has an ordinary open interval contained in \(E\), so \(E\) is open in \(\mathbb R\) and therefore also in \(Y\).