9.3. Limit Points¶
On real line, we usually take advantage of the notion of “closeness” in the form of distance to compute limits of sequences.
In a general topological space we don’t have a “distance function”.
We define the notion of limit point without resorting to distances.
We will also introduce the notion of connectedness.
In \(\mathbb{R}\), while the sets \([0,1]\cup[2,3]\) and \([4,6]\) both have length 2, but they are different in the sense of connectedness.
9.3.1. Limit Points and Closure¶
If \((X, \mathcal{T})\) is a topological space then it is usual to refer to the elements of the set \(X\) as points.
Definition
Let \(A\) be a subset of a topological space \((X, \mathcal{T})\). A point \(x \in X\) is said to be a limit point (or accumulation point or cluster point) of A if every open set, \(U\) containing \(x\) contains a point of \(A\) different from \(x\).
Example
\(X = \{a,b,c,d,e\}\)
\(\mathcal{T} = \{X, \phi, \{a\},\{c,d\},\{a,c,d\}, \{b,c,d,e\}\}\)
\(A = \{a,b,c\}\)
Note that \(A\) is neither an open set nor a closed set.
\(b, d, e\) are limit points of \(A\).
\(a,c\) are not limit points of \(A\).
The limit point need not belong to \(A\). \(d,e \notin A\).
Every point in \(A\), need not be a limit point. \(a,c\in A\) but they are not limit points.
Example
Let \((X, \mathcal{T})\) be a discrete space and \(A \subseteq X\).
\(A\) has no limit points since for each \(x \in X\), we have the singleton set \(\{x\} \in \mathcal{T}\) containing no point of \(A\) different from \(x\).
Example
Consider the subset \(A = [a,b) \subseteq \mathbb{R}\).
Every element in \([a,b)\) is a limit point of \(A\).
The point \(b\) is also a limit point of \(A\).
Example
A singleton set has no limit points.
How do we test whether a set is closed?
Proposition
Let \(A\) be a subset of a topological space \((X,\mathcal{T})\). Then \(A\) is closed in \((X,\mathcal{T})\) if and only if \(A\) contains all of its limit points.
Example
The set \([a,b)\) is not closed in \(\mathbb{R}\), since \(b\) is a limit point and \(b \notin [a,b)\).
The set \([a,b]\) is closed in \(\mathbb{R}\), since all limit points belong to the set itself.
The set \((a,b)\) is not closed in \(\mathbb{R}\), since \(a\) is a limit point and \(a \notin (a,b)\).
\([a,\infty)\) is a closed subset of \(\mathbb{R}\).
Proposition
Let \(A\) be a subset of a topological space \((X, \mathcal{T})\) and \(A'\) the set of all limit points of \(A\). Then \(A \cup A'\) is a closed set.
Definition
Let \(A\) be a subset of a topological space \((X, \mathcal{T})\). Then the set \(A \cup A'\) consisting of \(A\) and all its limit points is called the closure of \(A\) and is denoted by \(\overline{A}\).
Example
\(X = \{a,b,c,d,e\}\)
\(\mathcal{T} = \{X,\phi, \{a\}, \{c,d\}, \{a,c,d\} , \{b,c,d,e\} \}\)
The only limit point of \(\{b\}\) is \(e\)
\(\overline{\{b\}} = \{b,e\}\).
The limit points of \(\{a,c\}\) are \(b,d,e\)
Thus \(\overline{\{a,c\}} = X\).
Remark
Every closed set containing \(A\) must also contain the set \(A'\). So \(A \cup A' = \overline{A}\) is the smallest closed set containing \(A\). This implies that \(\overline{A}\) is the intersection of all closed sets containing \(A\).
Example
\(\overline{\mathbb{Q}} = \mathbb{R}\).
Definition
Let \(A\) be a subset of a topological space \((X, \mathcal{T})\). Then \(A\) is said to be dense in \(X\) or everywhere dense in \(X\) if \(\overline{A} = X\).
Example
Let \((X, \mathcal{T})\) be a discrete space. Then the only dense subset of \(X\) is \(X\) itself.
Each subset is its own closure since every subset is both open and closed.
Proposition
Let \(A\) be a subset of a topological space \((X, \mathcal{T})\). Then \(A\) is dense in \(X\) if and only if every non-empty open subset of \(X\) intersects \(A\) non-trivially (i.e. if \(U \in \mathcal{T}, U \neq \phi\), then \(A \cap U \neq \phi\))
Let \(S\) and \(T\) be non-empty subsets of a topological space \((X,\mathcal{T})\) with \(S \subseteq T\).
If \(p\) is a limit point of the set \(S\), then \(p\) is also a limit point of the set \(T\).
Thus \(\overline{S} \subseteq \overline{T}\).
If \(S\) is dense in \(X\), then \(T\) is dense in \(X\) too.
9.3.2. Neighborhoods¶
Definition
Let \((X, \mathcal{T})\) be a topological space, \(N\) a subset of \(X\) and \(p\) a point in \(N\). Then \(N\) is said to be a neighborhood of the point \(p\) if there exists an open set \(U\) such that \(p \in U \subseteq N\).
Example
The closed interval \([0,1]\) in \(\mathbb{R}^2\) is a neighborhood of the point \(\frac{1}{2}\) since \(\frac{1}{2} \in (\frac{1}{4}, \frac{3}{4}) \subseteq [0,1]\).
The interval \((0,1]\) in \(\mathbb{R}^2\) is a neighborhood of the point \(\frac{1}{4}\) since \(\frac{1}{4} \in (0, \frac{1}{2}) \subseteq (0,1]\). But \((0,1]\) is not a neighborhood of the point \(1\).
Example
If \((X, \mathcal{T})\) is a topological space and \(U \in \mathcal{T}\), then \(U\) is a neighborhood of every point \(p \in U\).
Every open interval \((a,b)\) is a neighborhood of every point it contains.
Let \(N\) be a neighborhood of a point \(p\). Then If \(S\) is any subset of \(X\) such that \(N \subseteq S\), then \(S\) is also a neighborhood of \(p\).
Proposition
Let \(A\) be a subset of a topological space \((X,\mathcal{T})\). A point \(x \in X\) is a limit point of \(A\) if and only if every neighborhood of \(x\) contains a point of \(A\) different from \(x\).
Corollary
Let \(A\) be a subset of a topological space \((X, \mathcal{T})\). Then the set \(A\) is closed if and only if for each \(x \in X\setminus A\), there is a neighborhood \(N\) of \(x\) such that \(N \subseteq X\setminus A\).
Corollary
Let \(U\) be a subset of a topological space \((X,\mathcal{T})\). Then \(U \in \mathcal{T}\) if and only if for each \(x \in U\) there exists a neighborhood \(N\) of \(x\) such that \(N \subseteq U\).
Corollary
Let \(U\) be a subset of a topological space \((X,\mathcal{T})\). Then \(U \in \mathcal{T}\) if and only if for each \(x \in U\) there exists a \(V \in \mathcal{T}\) such that \(x \in V \subseteq U\).
A set is open if and only if it contains an open set about each of its points.
Definition
A topological space \((X,\mathcal{T})\) is said to be separable if it has a dense subset which is countable.
Example
\(\mathbb{Q}\) is countable and a dense subset of \(\mathbb{R}\). Hence \(\mathbb{R}\) is separable.
Definition
Let \((X,\mathcal{T})\) be a topological space and \(A\) any subset of \(X\). The largest open set contained in \(A\) is called the interior of \(A\) and is denoted by \(\text{Int}(A)\).
\(\text{Int}([0,1]) = (0,1)\)
\(\text{Int}((3,4)) = (3,4)\)
\(\text{Int}({5}) = \phi\)
If \(A\) is open in \((X,\mathcal{T})\), then \(Int(A) = A\)
9.3.3. Connectedness¶
Let \(S\) be any set of real numbers.
If there is an element \(b \in S\) such that \(x \leq b \forall x \in S\), then \(b\) is called the greatest element of \(S\).
If \(S\) contains an element \(a\) such that \(a \leq x \forall x \in S\), then \(a\) is called the least element of \(S\).
A set \(S\) of real numbers is called bounded above if there exists a real number \(c\) such that \(x \leq c \forall x \in S\), and \(c\) is called an upper bound for \(S\).
Similarly the terms bounded below and lower bound are defined.
A set which is bounded above and bounded below is called bounded.
Least Upper Bound Axiom
Let \(S\) be a non-empty set of real numbers. If \(S\) is bounded above, then it has a least upper bound.
The least upper bound, a.k.a. the supremum of \(S\), denoted by \(\sup(S)\) may or may not belong to \(S\).
Any set of real numbers which is bounded below has a greatest lower bound which is also called the infimum and is denoted by \(\inf(S)\).
Lemma
Let \(S\) be a subset of \(\mathbb{R}\) which is bounded above and let \(p\) be the supremum of \(S\). If \(S\) is a closed subset of \(\mathbb{R}\), then \(p \in S\).
Proposition
Let \(T\) be a clopen set of \(\mathbb{R}\). Then either \(T = \mathbb{R}\) or \(T = \phi\).
Definition
Let \((X, \mathcal{T})\) be a topological space. Then it is said to be connected if the only clopen sets of \(X\) are \(X\) and \(\phi\).
Proposition
The topological space \(\mathbb{R}\) is connected.
Example
If \((X, \mathcal{T})\) is any discrete space with more than one element, then \((X, \mathcal{T})\) is not connected as each singleton set is clopen.
If \((X, \mathcal{T})\) is an indiscrete space, then it is connected as the only clopen sets are \(X\) and \(\phi\).
Remark
A topological space \((X, \mathcal{T})\) is not connected (that is disconnected) if and only if there are non-empty open sets \(A\) and \(B\) such that \(A \cap B = \phi\) and \(A \cup B = X\).
\(\mathbb{R}^2\) and in general \(\mathbb{R}^n\) are connected spaces.