TITLE: Axiom of choice in set theory
QUESTION [7 upvotes]: Just as the title stated, what is the main point of axiom of choice? I do not quite understand what is written in the axiom. The axiom that I know is:

Given any collection of non-empty sets, there exists a choice function such that $$f:I \rightarrow \bigcup_i{S_i}\quad f(i)\in S_i \quad\text{for all }\; i\in I.$$

REPLY [5 votes]: The axiom of choice states that when given a family of nonempty sets, we can choose an element from each set. Hence the name, the axiom of choice. 
It is useful because it allows us to control the behavior of infinite objects. It ensures that we can prove the existence of things which otherwise may require an infinitely long statement, for example a Hamel basis for every vector space. 
When writing a proof about an infinite collection of objects we will sometimes want to make some arbitrary choice amongst them and just "run with it". It might be possible to make that choice less arbitrary, for example if we talk about finite sets of real numbers you can always take the minimal element; or if you talk about continuous functions from a compact interval to $\mathbb R$, you can talk about the smallest point which attains the maximal value.
When you have a uniform way of making a choice, that is one "algorithm" which ensures that a choice is made, then you really have no need for the axiom of choice. The axiom of choice supplies this method when you cannot prove it exists otherwise.

Further reading:

Simple and intuitive example for Zorns Lemma
Making a choice function if $A_i$ are well-ordered for each $i$
Is there any motivation for Zorn's Lemma?
Where is the Axiom of choice used?
Does proving (second countable) $\Rightarrow$ (Lindelöf) require the axiom of choice?
Axiom of Choice Examples
A concrete example of a choice function
Ultra Filter and Axiom of Choice
Every Hilbert space has an orthonomal basis - using Zorn's Lemma
Is Banach-Alaoglu equivalent to AC?
Axiom of choice and compactness.
Is the axiom of choice needed to show that $a^2=a$?
Advantage of accepting non-measurable sets
Algebraic closure for $\mathbb{Q}$ or $\mathbb{F}_p$ without Choice?
A question concerning on the axiom of choice and Cauchy functional equation
Advantage of accepting the axiom of choice

Some of these threads discuss implications of the axiom of choice, or its common equivalents the well-ordering principle and Zorn's lemma. Some of them discuss where these principles are used and what may happen when they fail.