Real analysis exam 2

A sequence $(a_n)$ is Cauchy if, for every $\varepsilon>0$, there exists $N\in\mathbb N$ such that
$ m,n\ge N\implies |a_n-a_m|<\varepsilon. $

Every Cauchy sequence of real numbers converges, and every convergent sequence of real numbers is Cauchy.

Every bounded sequence of real numbers has a convergent subsequence.

A series is written as
$ \sum_{n=1}^{\infty}a_n. $

Its sequence of partial sums is
$ s_N=\sum_{n=1}^{N}a_n. $

The series converges if $(s_N)$ converges to a finite real number. Otherwise, it diverges.

If
$ \lim_{n\to\infty}a_n\ne0 $

or the limit does not exist, then $\sum a_n$ diverges.

If the limit equals zero, the test is inconclusive.

$ \sum_{n=0}^{\infty}ar^n $

Converges when $|r|<1$, with sum $a/(1-r)$. It diverges when $|r|\ge1$, except for the trivial zero series.

For nonnegative terms, if $0\le a_n\le b_n$:
- If $\sum b_n$ converges, then $\sum a_n$ converges.
- If $\sum a_n$ diverges, then $\sum b_n$ diverges.

$ \sum_{n=1}^{\infty}\frac1{n^p} $

- Converges if $p>1$.
- Diverges if $p\le1$.

A series whose terms alternate in sign, commonly $\sum(-1)^{n-1}b_n$ for $b_n\ge0$.

$\sum |a_n|$ converges.

$\sum a_n$ converges, but $\sum |a_n|$ diverges.

Every point in the set has a small open interval around it contained in the set.
For all $x \in O$, there exists $U_\varepsilon (x)\subseteq O$

The set contains all its limit points.

In $\mathbb R$, a set is compact exactly when it is closed and bounded.

A closed set with no isolated points.

Cannot be separated into two disjoint, nonempty sets that are open relative to the set.

Can be separated into two such sets.

Every open interval around the point contains a point of the set different from the point itself.

$x \in A \subseteq \mathbb(R) \iff \forall \varepsilon > 0$, $\exists y\in (U)_\varepsilon \cap A$ with $x\neq y$

A point of the set that has some open interval containing no other points of the set.

$\exists \varepsilon >0$ s.t. $U_\varepsilon (m)$ contains $m$ as the only elsement for the set $X$

Every open interval around the point intersects both the set and its complement.
$z \in A\subseteq \mathbb R \iff \forall \varepsilon >0$ s.t. $U_\varepsilon (z)$ contains $A , (A\subseteq U_\varepsilon (z))$ and $ A^\complement , (A^\complement \subseteq U_\varepsilon (z))$

A collection of open sets whose union contains the entire set being covered.

A subset of $\mathbb R$ is compact if and only if it is closed and bounded.