Applied And Computational Mathematics

Sunday, December 24, 2017

Limit Point & Closure



In Real analysis:

Let \(S\) \(\in\) \(\mathbb{R}\) be any set, and let \(x\) \(\in\) \(\mathbb{R}\) then: 
the point \(x\) is said to be limit point or a cluster point or an accumulation point or condense point of \(S\) if \(\forall \epsilon > 0\) , \( (x-\epsilon,x+\epsilon)\cap\) \(S \setminus {x} \neq \phi\)
           
In Topological Space: 

Let \(A\) be a subset of a topological space \((X,\tau)\). A point \(x \in X\) is said to be limit point of \(A\) if every open set, \(U\), containing \(x\) contains a point of \(A\) different from \(x\).  

Closed Sets: 
A Set \(S\) is said to be closed if one of these conditions achieved:  
  •  It's complement is open.
  • Or, it contains all its limit points.
Notions:
  1. The set of limit points of a set \(S\) is denoted by \(L(S)\)
  2. The subset \(X\) is a closed subset of itself.
  3. The Empty set is closed.
  4. Any Finite set is closed.
  5. A Subset of \(\mathbb{R}\) with no limit points is a closed set. ( See Example #6 ).
  6. In the metric space, the open disc \(\left \{ \left ( a,b \right ) \in R^{2} \mid a^{2}+b^{2} < 1 \right \}\) has the closed disc \(\left \{ \left ( a,b \right ) \in R^{2} \mid a^{2}+b^{2} \leqslant 1 \right \}\) as limit points.
Examples: 

Example #1: Consider the topological space \((X,\tau)\) where the set \(X = \left \{ a,b,c,d,e \right \}\), the topology \(\tau =\) \(\left \{X,\phi, \left \{ a\right \},  \left \{ c,d \right \},  \left \{ a,c,d \right \},  \left \{ b,c,d,e \right \}\right \}\), and \(A=\left \{ a,b,c \right \}\). Then \(b\), \(d\), and \(e\) are limit points of \(A\) but \(a\) and \(c\) are not limit points of \(A\). Explain ?  

Solution: 
-The set \( \left \{ a \right \} \) is open and contains no other points of A. Hence, \(a\) is not a limit point of \(A\) even though the set \( \left \{ a,c,d \right \} \) contains other points of \(A\), because not every open set containing \(a\) contains a point of \(A\) different from \(a\).  
-Now, \(b\) is a limit point of \(A\) because the only open sets containing \(b\) are \(X\) and \( \left\{b,c,d,e \right\} \) contains at other element of \(A\) which is \(c\). Thus, \(b\) is a limit point of \(A\).  
-Choosing \( \left \{c,d \right\} \) as an open set containing \(c\), will exclude \(c\) from being limit point of \(A\), because it contains no other point of \(A\) other than \(c\). Hence, \(c\) is not limit point of \(A\). 
-Moreover, an interesting fact that the two elements \(e\) and \(d\) are limit points of \(A\) even they are not in \(A\). And this is because every open set in \(\tau\) containing either \(d\) or \(e\) will contain point of \(A\) other than these points. Thus, \(d\) and \(e\) are limit points of \(A\). 

Example #2: Let \(S=\left (a,b  \right )\) and \(x \in \left (a,b \right ) \). Show that \(x\) is a limit point of \(S\). 

Solution: 
Let \( \epsilon > 0 \) and consider the interval \( \left (x-\epsilon,x+\epsilon \right ) \). Then, this interval will contain points of \( \left (a,b \right ) \) other than \(x\) .i.e \(\forall \epsilon > 0\) , \( (x-\epsilon,x+\epsilon)\cap\) \(S \setminus {x} \neq \phi\). In fact, it contains many infinite points other than \(x\). Similarly, we can show that \(a\) and \(b\) are also a limit point of the open interval \( \left (a,b \right )\) and the closed interval \( \left [ a,b\right ] \).
In General, \(L\left (\left (a,b \right )\right )\) = \( \left [ a,b\right ] \).

Example #3: Let \(S=\left (0,1 \right ) \cup \left \{ 2 \right \}\) , then show that:
  1. \(2\) is close to \(S\) 
  2. \(2\) is not limit point of \(S\).
  3. What is the closure of \(S\) ?
  4. Find the limit points of \(S\), .i.e \(L(S)\) ?
Solution: 
  1. It is obvious that \( \forall \epsilon > 0 \), \(\left \{ 2 \right \}\) \(\subseteq \) \(\left (2-\epsilon,2+\epsilon \right ) \cap S \) such that \(\left (2-\epsilon,2+\epsilon \right ) \cap S \) \(\neq\phi\).
  2. Applying the definition above, \( \forall \epsilon > 0 \),  \(\left (2-\epsilon,2+\epsilon \right ) \cap S \) \(\setminus \left \{ 2 \right \}=\phi\). Thus \(2\) is not a limit point of \(S\).
  3. To find the closure of \(S\) , .i.e \(\bar{S}\), we need to find the smallest closed interval that contain \(S\). In tacit, the set \( \left [ 0,1\right ] \cup \left \{ 2 \right \} \) is the closure of \(S\).
  4. We can show that the points \(a\) and \(b\) are both limit points of \(S\). Also, any point belong to the open interval \(\left(a,b\right ) \) is limit point of \(S\). Thus, all limit points of \(S\) belong to the closed interval \(\left[a,b \right]\). Hence, \(L(S)=\left[a,b \right]\).
Example #4: Show that the set of all integer numbers has no limit points, .i.e \(L\left( \mathbb{Z} \right) = \phi\).

Solution:
Let \( x\in\mathbb{R} \). We say that \(x\) is limit point of \( \mathbb{Z} \) iff \( \forall \epsilon > 0 \) the interval \( \left(x-\epsilon,x+\epsilon \right ) \) contains points of \( \mathbb{Z} \) other than \( x \). But \( \forall x \in \mathbb{R} \), \( \exists n \in \mathbb{Z} \) such that \( n-1 \leq x < n \). so, if \(\ epsilon < min \left ( \left | x-n+1 \right |,\left | x-n \right | \right ) \) then the interval \( \left (x-\epsilon , x+\epsilon \right ) \) will contain no integer. Therefore, \( \left (x-\epsilon , x+\epsilon \right ) \cap \mathbb{Z}\setminus \left \{ x \right \}= \phi \). Thus, no real number can be limit point of \(\mathbb{Z}\).  
In analogy with \(\mathbb{N}\), we can show that \(L\left( \mathbb{N} \right) = \phi\) either.

Example #5: Show that the set of all rational number has infinitely many limit points, .i.e \(L\left(\mathbb{Q} \right) = \mathbb{R}\).

Solution:
By the definition, \( \forall \epsilon > 0 \), the interval \(  \left (x-\epsilon , x+\epsilon \right ) \) contains infinitely many points of \( \mathbb{Q} \) other than \(x\). Hence,  \( \left (x-\epsilon , x+\epsilon \right ) \cap \mathbb{Q}\setminus \left \{ x \right \} \neq \phi \). Therefore, every real number is limit point of \( \mathbb{Q}\). 

Example #6: Show (in term of limit points) that the set of all natural number \(\mathbb{N}\) is closed set.

Solution:
To prove that \(\mathbb{N}\) is closed, we have to show that it contains all its limit points. And we knew that \(L\left( \mathbb{N} \right) = \phi\). Now, since \( \phi \subset \mathbb{N}\) then \(\mathbb{N}\) contains all of limit point. i.e, \(\phi \cap \mathbb{N} \setminus \phi \neq \phi\). 

Example #7: Let \( S=\left \{ \frac{1}{n}:n \in \mathbb{Z}^{+} \right \} \). Show that \( L \left ( S \right ) = \left \{ 0 \right \} \).

Solution:
Implement the following set on the real line to have the following sketch: 


The only limit point appear here is the \(0\) point, since \( \forall \epsilon > 0 \), the interval \( \left (0-\epsilon , 0+\epsilon \right ) \) contains points of \(S\) other than the zero point itself. Therefore, using the difinition of the limit point, we can say that  \( \left (0-\epsilon , 0+\epsilon \right ) \cap S \setminus 0 \neq \phi\). Hence the only limit point of \(S\) is the Zero point. 





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