In a metric space $\Omega$ we shall define open sets. Given sets $A,B\subset \Omega$ and
the distance function $d(x,y)$ .
We call a set $B(x,\delta)\in\Omega$ a open ball centered at a point $x$ which has the radius $\delta$ if, for any element $x\in\Omega$ and a real number $\delta>0$ , $B$ is a set $\left\{ y | d(x,y)<\delta \right\}$ .
Using a open ball, a set $A$ is called a neighbourhood of $x$ if $A$ has a subset $B(x,\delta)$ .
Open sets in $\Omega$ is the set in which all elements has a neighbourhood of the element
and which contains the neighbourhood.
Namely, a set $O$ is called open if, for any point $x\in O$ . there is a real number $\delta$
such that a set $\left\{ y | d(x,y)<\delta \right\}$ belongs to $O$ .
$R^n$ is open and the empty set $\phi$ is too. (After this, you will find that these two sets are also closed sets. )
Open sets are the base of topological space.
2014/03/08
direct product
If two sets $A$ and $B$ are given, we are able to make ordered combinations of each element of the set. Namely if $a\in A$ and $b\in B$ , one of ordered combinations is $(a,b)$ . These are called direct product. We will show all combinations $A\times B$ . Hence,
$A\times B=\left\{ (a,b) | a\in A, b\in B \right\}$
You have to note that direct product is ordered. Therefore, $(a,b)$ is not equal to $(b,a)$ . Because two sets are arbitrary, $A=B$ is allowed. If $A=B=\mathbb{R}$ , $(a,b)$ means the point of Cartesian coordinates of $\mathbb{R}\times \mathbb{R}=\mathbb{R}^2$ . Then, elements $a$ and $b$ become coordinate axes.
Direct product is expanded over two sets. Given n sets, it means n-dimensional space. You must have known $\mathbb{R}^3$ very well. Having already proved, you may still remember that, for $C=\left\{ 0,1 \right\}$ ,
$C^{\infty}=\left\{ 0,1 \right\}\times \left\{ 0,1 \right\}\times \cdots$
is uncountable.
$A\times B=\left\{ (a,b) | a\in A, b\in B \right\}$
You have to note that direct product is ordered. Therefore, $(a,b)$ is not equal to $(b,a)$ . Because two sets are arbitrary, $A=B$ is allowed. If $A=B=\mathbb{R}$ , $(a,b)$ means the point of Cartesian coordinates of $\mathbb{R}\times \mathbb{R}=\mathbb{R}^2$ . Then, elements $a$ and $b$ become coordinate axes.
Direct product is expanded over two sets. Given n sets, it means n-dimensional space. You must have known $\mathbb{R}^3$ very well. Having already proved, you may still remember that, for $C=\left\{ 0,1 \right\}$ ,
$C^{\infty}=\left\{ 0,1 \right\}\times \left\{ 0,1 \right\}\times \cdots$
is uncountable.