infimum implies existence of a sequence

Question

I am following 'Introductory Functional Analysis' by Kreyszig. Theorem 3.3-1 says that

Let $X$ be an inner product space and $M\neq\phi$ a convex subset which is complete (in the metric induced by inner-product). Then for every given $x\in X$ there exists a unique $y\in M$ such that $$\delta=\inf_{y'\in M}||x-y'||=||x-y||.$$

In the proof of this theorem, he has written that

By the definition of infimum there is a sequence $(y_n)$ in $M$ such that $\delta_n$ approaches $\delta$ where $\delta_n=||x-y_n||.$

I don't understand this statement because definition of infimum does not say anything about the existence of a sequence. For example, I can have a set {1} which has infimum equal to 1, but there is no sequence which converges to 1. So what does he mean in this proof? Am I missing something here?

Answer 1

Given a set $A$ with infimum $a$, choose an element $a_1 \in A$ satisfying $a_1 < a + 1$; this exists by the definition of the infimum. Choose an element $a_2 \in A$ satisfying $a_2 < a + 1/2$; this exists by the definition of the infimum. Rinse and repeat.

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In the case of a set like $\{1\}$, the sequence is constant. This is not a problem.