Proof of Topology and Topological Space Through Axioms

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Topology and Topological Space

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Proof of Topology and Topological Space Through Axioms

 

The Three Axioms

(i) $X$ and the empty set $ϕ$ belongs to $T$
(ii) $X$ the union of any finite or infinite number of sets $T$ belongs to $T$ and
(iii) $X$ the intersection of any two sets $T$ belongs to $T$

The pair $(X,T)$ is called a topological space. We can also tell:
a set $X$for which a topology $T$ has been specified is called a topological space.

Among the above mentioned three axioms, we will first start with (i)

The first axiom:

(i) $X$ and the empty set $ϕ$ belongs to $T$ can be re-written as $ϕ,X∈T$. If we look closely, then any finite intersection $U_{1}∩U_{2}… U_{n}$ of open sets is open.

 

What do We Get?

 

If $X $is a topological space with topology $T$, we say that the subset $U$ of $X$ is an open set of $X$ if $U$ belongs to this collection $T $ (James R. Munkres)

From here we will take a twofold approach:

  1. Showing through visual, which one has a topology and which has not.
  2. Showing through sets (using numbers) which one has a topology and which has not.

 

Showing Through Visuals

 

Let $M$ be a three-element set $M={a,b,c}$

 

Figure 19 - Topology of sets
Figure 19 – Topology of sets

 

Figure 19 shows the topology in which open sets are $X$, $ϕ$ $\{ x,y \}$,$\{y\}$ and $\{ y,z \}$

 

Figure 20- Topology of open sets
Figure 20- Topology of open sets

 

Figure 20 contains only $X$ and $ϕ$

 

Figure 21- Non topological sets
Figure 21- Non-topological sets

 

Figure 21 shows that neither of the collections indicates is a topology

 

Showing Through Sets

 

Let $X = \{a,b,c,d,e,f\}$ and
$T_1= X,ϕ,\{a\},\{c,d\},\{a,c,d\},\{b,c,d,e,f\}$ then
$T_1 $is a topology as it satisfies:

(i) $X$ and the empty set $ϕ$ belongs to $T$
(ii) $X$ the union of any finite or infinite number of sets $T$ belongs to $T$ and
(iii) $X$ the intersection of any two sets $T$ belongs $T$

The pair $(X,T)$ is called a topological space.

Let $X$=$\{a,b,c,d,e\}$ and

$T_2$ = $\{X,\phi\},\{a\},\{c,d\},\{a,c,e\},\{b,c,d\}\}$ then

$T_2$ is not topology as the union:

 

$\{c,d\}∪\{a,c,e\}=\{a,c,d,e\}$

 

Two members of $T_2$ does not belong to $T_2$ hence, $T_2$ does not satisfy:

$X$ the union of any finite or infinite number of sets $T$ belongs to $T$

 

Let us consider $N$ as a set that has all numbers which are natural and
$T_3=\{ N,ϕ, all finite subsets of N\}$ then

$T_3$ not a topology on $N$, as the infinite union

$\{2\}$ ∪$\{3\}$ $∪$…$∪$$\{n\}…$ = $\{2,3,…n…\}$ of members $T_3$ does not belong to $T_3$,

that is $T_3$ does not have the property: 

$X$ the union of any finite or infinite number of sets $T$ belongs to $T$

 

Let $X = \{a,b,c,d,e,f\}$ and

$T_4$ = $\{X,\phi\},\{a\},\{f\},\{a,f\},\{a,c,f\},\{b,c,d,e,f\}\}$ then

$T_4$ is not topology on $X$ as the intersection:

$\{a,c,f\} ∩\{b,c,d,e,f\} = \{c,f\}$

two members of $T_4$ does not belong to $T_4$ – $T_4$ does not satisfy:

$X$ the intersection of any two sets $T$ belongs $T$

This concludes our discussion on the basic concepts of topology.

 

See Also | More Topology Articles


Further References:

  • Topology – James R. Munkres
  • Topology without tears – Sidney A.Morris

Article Advisor:

  • Richard Sot (PhD. Mathematics, University of Rochester, Rochester, NY)

Website references:

  1. https://mathworld.wolfram.com
  2. https://www.wikipedia.org/
  3. https://math.stackexchange.com/
  4. https://www.researchgate.net/publication/343635292
  5. https://www.sciencedirect.com/topics/chemistry/space-topological

 

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