Let \(X\subset\mathbb R^n\) be closed and let \(r>0\). Define \[ Y=\{y\in\mathbb R^n:|x-y|=r\text{ for some }x\in X\}. \] Show that \(Y\) is closed.
Solution
1. Let \((y_k)\) be a sequence in \(Y\) converging to\[ y\in\RR^n. \]For each \(k\), choose \(x_k\in X\) such that\[ |x_k-y_k|=r. \]
This is possible by the definition of \(Y\).
2. The sequence \((x_k)\) is bounded.
Since \(y_k\to y\), the sequence \((y_k)\) is bounded. Hence there is \(M>0\) such that \[ |y_k|\leq M \] for all \(k\). By the triangle inequality, \[ |x_k| \leq |x_k-y_k|+|y_k| \leq r+M. \] Thus \((x_k)\) is bounded.
3. Some subsequence \((x_{k_j})\) converges to a point \(x\in X\).
By step 2 and the Bolzano–Weierstrass theorem in \(\RR^n\), \((x_k)\) has a convergent subsequence \[ x_{k_j}\longrightarrow x. \] Every \(x_{k_j}\) lies in the closed set \(X\), so its limit \(x\) also lies in \(X\).
4. The limit point \(y\) belongs to \(Y\).
Along the same subsequence, \[ y_{k_j}\longrightarrow y. \] By continuity of the Euclidean norm, \[ |x-y| = \lim_{j\to\infty}|x_{k_j}-y_{k_j}| = r. \] Since \(x\in X\) by step 3, the definition of \(Y\) gives \(y\in Y\).
5. Therefore\[ \boxed{Y\text{ is closed}}. \]
Steps <1>1–<1>4 show that every convergent sequence in \(Y\) has its limit in \(Y\). In the metric space \(\RR^n\), this is equivalent to closedness.
6. Q.E.D.
Step 5 is the required conclusion.