# Mathematics Prelims

## July 15, 2008

### Some Basic Topology, Part I

Filed under: Analysis,Functional Analysis,Topology — cjohnson @ 12:16 pm

A topology on a set $X$ is a collection of subsets of $X$, which we’ll call $\tau$, such that the following are satisfied

• $\emptyset, X \in \tau$
• If $\Lambda$ is some index set where $\forall \lambda \in \Lambda : U_\lambda \in \tau$, then $\bigcup_{\lambda \in \Lambda} U_\lambda \in \tau$
• If $U_1, ..., U_n \in \tau$ for any $n \in \mathbb{N}$, then $\bigcap_{i=1}^n U_i \in \tau$.

If $(X, d)$ is a metric space, then we can generate a topology $\tau$ on $X$ by considering the collection of all “open balls” in $X$.

Given any point $x$ in a metric space $(X, d)$ and an $r > 0$, the open ball of radius $r$ centered at $x$, denoted $B_r(x)$, is the set of all points in the space that are less than distance $r$ from $x$.  That is, $B_r(x) = \{ y \in X : d(x, y) < r \}$.  The closed ball of radius $r$ centered at $x$ is defined similarly: $\overline{B_r}(x) = \{ y \in X : d(x, y) \leq r \}$.  Finally, the sphere of radius $r$ centered at $x$ consists of those points that are precisely distance $r$ from $x$: $S_r(x) = \{ y \in X : d(x, y) = r \}$.

If $M$ is some subset of $X$ and $m \in M$, then we say that $m$ is an interior point of $M$ if there exists an $r > 0$ such that $B_r(m)$ is contained in $M$.  If every point of a set is an interior point, then we say that the set is open.  For any set, open or not, we define the interior of the set to be all of the sets interior points.

Note that the interior of a set is the largest open set contained in that set.  For example, suppose $M \subseteq X$ and let $M^\circ$ be the interior of $M$ (there isn’t, as far as I know, any real standard notation for the interior of a set).  By the definition of the interior, $M^\circ \subseteq M$ and each point in $M^\circ$ can have an open ball placed around it that stays within the set, so $M^\circ$ is an open subset of $M$.  Suppose $I$ is also an open subset contained in $M$ and that $M^\circ \subset I \subseteq M$.  This means that there must exist an $x \in I$ and a corresponding $r > 0$ such that $B_r(x) \subseteq I \subseteq M$.  However, by assumption, $M^\circ$ is the interior of $M$ so it must contain all such points, and no such $I$ can exist, thus $M^\circ$ is the largest open subset contain in $M$

The point $x$ is called a limit point (some authors call it an accumulation point) of $M \subseteq X$ if for every $r > 0$ there exists a $m \in B_r(x)$ such that $m \in M$ and $x \neq m$.  Note that a limit point of a set need not be an element of that set.  For instance, on the real line with the standard metric, 1 and 0 are both limit points of (0, 1) but neither is in the set.  (As the name “limit point” implies, we can construct a sequence in $M$ whose limit is $x$, but we’ll have to wait until we define the limit of a sequence later before we can show that.)

If $M \subseteq X$ and the complement of $M$ in $X$, $M^c = X \setminus M$, is open, then we say that $M$ is closed.  The closure of a set $M$, denoted $\overline{M}$, is the set $M$ along with all of its limit points.  This forms the smallest closed set that contains $M$.

Note that a set (unlike a door … bad joke) may be open, closed, both, or neither.  On the real line, for instance, the interval [0, 1) is neither open nor closed and the entire space, $\mathbb{R}$ is both open and closed.  In any topology, the empty set, $\emptyset$ and the entire space, $X$, are always both open and closed.  They are both open since, by the definition of a topology, they must be an element in the topology.  They are closed because they are complements of one another and the complement of an open set is a closed set.  In the discrete topology (where each subset of the space is an element of the topology) all sets are both open and closed.  (I’m not going to discuss it right now, but notice that the discrete topology is generated by the discrete metric.  To show that we’d need to talk about the basis of a topology, which we’re not going to do right now.)