> [!theorem] > > Let $\cx = \real^n$, then > $ > S(\cx) = \bracs{x \in \cx: \norm{x} = 1} > $ > with the [[Standard Topology|standard topology]] is [[Compactness|compact]]. > > *Proof*. Since $S(\cx)$ is the intersection of two closed sets $\bracs{x \in \cx: \norm{x} \ge 1}$ and $\norm{x \in \cx: \norm{x} \le 1}$, $S$ is closed. As $\real^n$ is [[Complete Metric Space|complete]], $S(\cx)$ is also complete. > > Now, let $\varepsilon > 0$, take balls centred at > $ > \bracs{\vec{x}: x_i = k_i\varepsilon, k_i\varepsilon \le 1 } > $ > then for any $\vec{y} = (y_1, \cdots, y_n)$ choose $k_i$ such that $y_n \in \varepsilon[k_i, k_{i + 1}]$, then $\vec{y} \in B(\varepsilon(k_1, \cdots, k_n), \varepsilon)$. > > Since the above collection of points is bounded by $(4/\varepsilon)^n$, it is a finite cover, and $S(\cx)$ is totally bounded. Therefore $S(\cx)$ is compact. > [!theorem] > > Let $\cx \in \complex^n$, then $S(\cx)$ as defined above is also compact. > > *Proof*. Let $\cy = \real^{2n}$ and $\phi: \cy \to \cx$ defined by > $ > (x_1, x_2, \cdots, x_{2n - 1}, x_{2n}) \mapsto (x_1 + x_2i, \cdots, x_{2n - 1} + x_{2n}i) > $ > then $\phi$ is bounded and therefore continuous. Since $S(\cx)$ is the continuous image of the compact set $S(\cy)$, $S(\cy)$ is also compact.