Closure of set of vectors with norm 1 in $\mathbb{C}^n$

Question

In the proof for existence of SVD, it always says - Due to compactness, we can always find a vector $v_{1} \in \mathbb{C}^n$ such that $A\,v_{1} = \sigma_{1} \, u_{1}$. Another post explained what is meant in the proof when they say compactness. However, I don't understand why the set $\left\{x \in \mathbb{C}^n \, | \, ||x|| = 1 \right\}$ is closed. I understand that it's bounded and that I need the set to be closed and bounded for compactness since we're in finite dimensions.

Answer 1

The Euclidean norm $\|\cdot\|:\mathbb C^n\to[0,\infty)$ is a continuous function, because if $(x_n)_{n\in\mathbb N}$ is a sequence in $\mathbb C^n$ converging to $x\in\mathbb C^n$, then the sequence of real numbers $(\|x_n\|)_{n\in\mathbb N}$ converges to the real number $\|x\|$. To see this, note that $$\big|\|x_n\|-\|x\|\big|\leq\|x_n-x\|\to 0.$$ The first inequality can be proved using the triangle inequality and the second convergence follows from the definition of what it means that $(x_n)_{n\in\mathbb N}$ converges to $x$ in $\mathbb C^n$.

Now, given that $\|\cdot\|$ is a continuous function, note that the preimage of the singleton $\{1\}$, which is a closed set in $\mathbb R$, is precisely the set $\{x \in \mathbb{C}^n \,|\, \|x\| = 1 \}$. The result follows from the fact that the preimage of a closed set under a continuous function is closed.

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An alternative proof: suppose that $x^*$ is in the closure of $\{x \in \mathbb{C}^n \,|\, \|x\| = 1 \}$. This means that there exists a sequence $(x_n)_{n\in\mathbb N}$ in the set such that $x_n\to x^*$. Now, one has that $\|x_n\|=1$ for each $n\in\mathbb N$. Therefore, $$\big|\|x^*\|-1\big|\leq\big|\|x^*\|-\|x_n\|\big|+\big|\underbrace{\|x_n\|-1}_{=0}\big|\leq\|x^*-x_n\|\to 0.$$ It follows that $\|x^*\|=1$, so that $x^*\in\{x \in \mathbb{C}^n \,|\, \|x\| = 1 \}$. One can now conclude that the closure of $\{x \in \mathbb{C}^n \,|\, \|x\| = 1 \}$ is included in the set $\{x \in \mathbb{C}^n \,|\, \|x\| = 1 \}$ itself, which means that this set is closed.