Open sets are a most important and basic concept. We will also understand it by following explanations.
Giving a metric space $\Omega$ and subsets $A,B\subset \Omega$ .
If $A,B\ne \phi$ , $A\cap B=\phi$ and both $A$ and $B$ are open sets,
$S=A\cup B$ is not a connected space (or set).
$S$ is called a disconnected space (or set).
A connected set means we can not divide it two open sets which are not empty
and of which the intersection is empty. If $S$ is connected, $A$ or $B$ should be empty.
A trivial example of a connected space is a set which consists of a point and the empty set.
Namely, a point is the connected space.
As it is a obviously definite thing that the line of real numbers is connected,
we get to a following basic proposition.
A connected set on the real number line is a interval. (a set is not empty and the interval may be open or closed. )
2014/04/19
some definitions related to open sets 2 (isolated and limit points)
Given a metric space $\Omega$ , a open set $A\subset \Omega$ and a member $a\in A$ .
We say a point $a$ is isolated, if, for any $\delta>0$ , a intersection of the open ball
$B(a,\delta )$ and $A$ is $a$ . It means there is not any point of $A$ except $a$
in a neighborhood of $a$ .
If any neighborhoods of $a$ has infinite points of $A$ , $a$ is called a accumulation point
or a limit point.
These will be understood easily. However, these are important definitions in the topology.
Please try to assume relations of various kinds of a set.
We say a point $a$ is isolated, if, for any $\delta>0$ , a intersection of the open ball
$B(a,\delta )$ and $A$ is $a$ . It means there is not any point of $A$ except $a$
in a neighborhood of $a$ .
If any neighborhoods of $a$ has infinite points of $A$ , $a$ is called a accumulation point
or a limit point.
These will be understood easily. However, these are important definitions in the topology.
Please try to assume relations of various kinds of a set.
2014/04/12
some definitions related to open sets
We shall introduce some definitions related to open sets. Here is a metric space $\Omega$ and $A,B\subset\Omega$ .
At first, the closed set means the complementary set of a open set. If $A$ is a open set,
$X=A^c$ is a closed set. Hence, $\mathbb{R}$ and the empty set $\phi$ are both closed sets.
Then $\mathbb{R}$ and the empty set $\phi$ become open sets and closed sets, too.
In a general metric space we have to accept sets which have these two properties at a same time.
There are sets which are not open and closed. You may remember half open intervals.
The interior $X$ of a set $A$ is the maximum open set of the set $A$ . Of course, if $A$ is a open set, $X=A$ . In other words, $X$ is the open set including all open sets which belong to $A$ . We often write just like $X=\cup\left\{ Y\subset A | Y \mbox{ is open} \right\}$ .
The closure $X$ of a set $A$ is the minimum closed set of the set $A$ . If $A$ is a closed set, $X=A$ . Correctly, $X$ is the minimum closed set which includes the set $A$ . We also often write $X=\cap\left\{ Y\supset A | Y \mbox{ is closed} \right\}$ .
The boundary $X$ of a set $A$ is the set whose elements are the closure minus the interior.
Namely, the intersections of the neighborhoods of any elements in the boundary $X$ and $A$ is not empty, and the intersections of the neighborhood and the complementary of $A$ is not empty, too.
These must be the most familiar definitions to you. However, you have to note that these are only based upon the definition of open sets.
At first, the closed set means the complementary set of a open set. If $A$ is a open set,
$X=A^c$ is a closed set. Hence, $\mathbb{R}$ and the empty set $\phi$ are both closed sets.
Then $\mathbb{R}$ and the empty set $\phi$ become open sets and closed sets, too.
In a general metric space we have to accept sets which have these two properties at a same time.
There are sets which are not open and closed. You may remember half open intervals.
The interior $X$ of a set $A$ is the maximum open set of the set $A$ . Of course, if $A$ is a open set, $X=A$ . In other words, $X$ is the open set including all open sets which belong to $A$ . We often write just like $X=\cup\left\{ Y\subset A | Y \mbox{ is open} \right\}$ .
The closure $X$ of a set $A$ is the minimum closed set of the set $A$ . If $A$ is a closed set, $X=A$ . Correctly, $X$ is the minimum closed set which includes the set $A$ . We also often write $X=\cap\left\{ Y\supset A | Y \mbox{ is closed} \right\}$ .
The boundary $X$ of a set $A$ is the set whose elements are the closure minus the interior.
Namely, the intersections of the neighborhoods of any elements in the boundary $X$ and $A$ is not empty, and the intersections of the neighborhood and the complementary of $A$ is not empty, too.
These must be the most familiar definitions to you. However, you have to note that these are only based upon the definition of open sets.
2014/04/05
open sets 2 (the intesection is open)
We will give a general proof by which, given open sets $O_1,O_2$ in a metric space $\Omega$ ,
the intersection of $O_1$ and $O_2$ is open.
A intersection of a finite number of open sets is open. Namely,
If $O_1,O_2,\cdots , O_k\in\Omega$ , $O_1\cap O_2,\cap \cdots \cap O_k\in\Omega$ is open.
If a element $e$ is in $\cap_{i=1}^k O_i$ , as $e\in O_i$ for all $i=1,\cdots ,k$ ,
there is a $\delta_i>0$ such that open ball $B(e, \delta_i)$ is in $O_i$ .
We choose $\delta=\min(\delta_1,\cdots\delta_k)$ .
Then, since $B(e, \delta)\subset B(e, \delta_i)$ for all $i=1,\cdots ,k$ ,
$B(e, \delta)\subset \cap_{i=1}^k O_i$ .
Thus, we obtain the result which we want. We should remember that a set $O$ is open
if and only if, for any element $e\in O$ , there is a $\delta>0$ such that the open ball $B(e,\delta)\subset O$ .
the intersection of $O_1$ and $O_2$ is open.
A intersection of a finite number of open sets is open. Namely,
If $O_1,O_2,\cdots , O_k\in\Omega$ , $O_1\cap O_2,\cap \cdots \cap O_k\in\Omega$ is open.
If a element $e$ is in $\cap_{i=1}^k O_i$ , as $e\in O_i$ for all $i=1,\cdots ,k$ ,
there is a $\delta_i>0$ such that open ball $B(e, \delta_i)$ is in $O_i$ .
We choose $\delta=\min(\delta_1,\cdots\delta_k)$ .
Then, since $B(e, \delta)\subset B(e, \delta_i)$ for all $i=1,\cdots ,k$ ,
$B(e, \delta)\subset \cap_{i=1}^k O_i$ .
Thus, we obtain the result which we want. We should remember that a set $O$ is open
if and only if, for any element $e\in O$ , there is a $\delta>0$ such that the open ball $B(e,\delta)\subset O$ .