I am having a little trouble understanding the definition of a complete set, which is the following : a metric space is said to be complete if every fundamental sequence converges in the space X.
Would this mean that the distance between all members of the set has to be greater than or equal to 0 and also has to be less than epsilon?
Am I totally off track?
Thanks for any help!!
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$\begingroup$We say a metric space is complete if every Cauchy sequence converges. A sequence is Cauchy if eventually the terms in the sequence get arbitrarily close to each other.
Here is my go-to example for thinking about a complete metric space.
The metric space $\mathbb{Q}$ of rationals is not complete. Why? Because we can find a sequence of rational numbers that is Cauchy but does not converge. It's actually really easy to do!
Think about the decimal expansion of $\pi = 3.14159 \dots$.
Let $x_{1} = 3$, $x_{2} = 3.1$, $x_{3} = 3.14$, $x_{4} = 3.141$, $x_{5} = 3.1415$, $x_{6} = 3.14159$, $\dots$.
Each term in this sequence is clearly a rational number, and this sequence is Cauchy (verifying this would be a good exercise for you to do).
Now, this sequence converges to $\pi$, which is also a good exercise for you to verify. But $\pi$ is not in $\mathbb{Q}$, since it is irrational. So we found a Cauchy sequence in $\mathbb{Q}$ that doesn't converge, which means $\mathbb{Q}$ is not complete.
You can also do the same thing to prove $\mathbb{R} - \mathbb{Q}$, the set of irrationals, is not complete. Let $x_{n} = 1 + \frac{\sqrt{2}}{n}$. This is a Cauchy sequence (check this!) and it converges to $1$ (check this!) but $1$ is rational, not irrational, so the sequence does not converge in $\mathbb{R} - \mathbb{Q}$.
$\endgroup$ $\begingroup$Yes, you're off track. By definition, distances in a metric space are greater than or equal to $0$. There is no "epsilon" that all distances have to be less than.
"Fundamental sequence" is more usually called a Cauchy sequence.
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