Mechanics of Continua and Structures
Definition (Archimedian Property) Given any positive $\epsilon>0$ there exists a natural number $n$ such that $1/n<\epsilon$.
Theorem The set of real numbers has the Archimedian property.
Corrollary Given any real number $x$ there exists a natural number $n$ such that $x<n$.
Proof If $x<0$ then choose $n=1$. If $x>0$ then $\epsilon:=1/x>0$ and there exits an $n$ such $1/n<\epsilon$. Or equivalently $1/n<1/x$. Multiplying the last inequality by $x$, and noting that both $x$ and $n$ are positive real numbers, we get $x<n$.
Definition A sequence ${x_{k}}\in (M,d)$ converges to $L\in M$ or has a limit $L\in M$ if for any $\epsilon>0$ there exists a $N\in \mathbb{N}$ such that $d(L,x_{k})<\epsilon$ $\forall$ $k\ge N$
For the above definition to make sense, $L$ should belong to $(M,d)$; otherwise the quantity $d(x_{k},L)$ does not make any sense. For example, consider the metric space $M~=~[0,1]$ . The sequence ${x_{k}}\in M, \, x_{k}=\frac{1}{k},\, k\in \mathbb{N}$ converges to $0\in M$. However, if we restrict $M=(0,1]$, then the limit point $0$ no longer lies in $M$ and hence it makes no meaning to talk about it, since the sequence does not converge and has no limit.
Definition (Cauchy Sequences) In a metric space $(M,d)$ a sequence ${x_{k}}$ is cauchy for every $\epsilon > 0$ there exists a $N\in\mathbb{N}$ such that $d(x_{m},x_{n})<\epsilon\, \forall \, n,m>N$
Example
The sequence $(1/k)_{k=1,2,\ldots}$ is Cauchy.
Proof From the Archimedian property of real numbers, we know that for any given $\epsilon>0$ there exits an $N\in \mathbb{N}$ such that $1/N<\epsilon/2$. Let $m$, $n\in \mathbb{N}$ be greater than $N$. Then we have from the triangle inequality that
\(\begin{align} \left|\frac{1}{m}-\frac{1}{n}\right|&\le \left|\frac{1}{m}\right|+\left|\frac{1}{n}\right|\\ < \frac{1}{m}+\frac{1}{n} \label{eq:mn} \end{align}\) Note that the second step simply follows from the fact that both $1/m$ and $1/n$ are positive real numbers.
Since $m,~n> N$, both $1/m$ and $1/n$ are less than $1/N$. Therefore, it follows from $\eqref{eq:mn}$ that
\[\begin{align} \left|\frac{1}{m}-\frac{1}{n}\right| &<\frac{2}{N}\\ &<\epsilon\\ \hfill \square \end{align}\]Definition (Cauchy Completeness) A metric space $(M,d)$ is called complete if every cauchy sequence in $M$ converges to a point in $M$
Remark For $M=(0,1]$ there is a sequence $(1/k)$ that is Cauchy but that does not converge to a point in $\mathcal{M}$. Hence the metric space $M=]0,1]$ is incomplete. There may be millions or in math language uncoutable infinite caucy sequences that converge. However, the existence of just one cauchy sequence that does not converge is sufficient to show that the space is not complete.
\[\begin{equation*} \mbox{Limit Points}=\mbox{Accumulation Pt.s}+\mbox{Isolated Pt.s} \end{equation*}\]There is general methodology taught by Prof. Shoen. We can complete this space by appending to it the limits of all the cauchy sequences by which all cauchy sequences will converge. To append the limits we need them to exist. That is to say, if $M=]0,1]$ then the the limit point of the sequence $x_{k}=1/k$ which is $0$ does not exist. (Remember: for us, our metric space is our universe.) However, we can embed this metric space in a new metric space called $\mathbb{R}^{1}$ isometrically, i.e, the distance between the points in $]0,1]$ remain the same. Now, however, the point $0$ exists. We complete our old metric space by taking its closure $\overline{M}=[0,1]$. Now our new metric space $\overline{M}=[0,1]$ is complete (or cauchy complete).
A totally bounded set is bounded but not the converse.
In $\mathbb{R}^{n}$ $\textit{boundedness}$ and $\textit{totally boundedness}$ are equivalent. For example, $\mathbb{Q}\cap[0,1]$ is totally bounded.