1.3 Open Sets
Definition 1.3.1. Let \(A\) be a subset of \(\mathbb {R}\). Then a point \(\, x\in \mathbb {R}\,\) is said to be an interior point of \(A\) if \(\, \exists \, \varepsilon > 0\, \ni \, \big ( x - \varepsilon \, ,\, x + \varepsilon \big ) \subset A\).
i.e \(x\) is an interior point of \(A\) if \(\, \exists \, \varepsilon > 0\, \ni \, \begin {vmatrix} y - x\\ \end {vmatrix} < \varepsilon \, \implies \, y \in A\).
The set of interior points of \(A\), denoted by \(A^0\) is called the interior of \(A\).
Note.
- 1.
- \(A^0\) is the set of all points of \(A\) for which \(A\) is a neighborhood.
- 2.
- \(A^0 \subset A\).
Recall that a real number \(\, x \in \overline {A} \, \iff \, \forall \, \varepsilon > 0\, \exists \, y \in A\ni \, \begin {vmatrix} y - x\\ \end {vmatrix} < \varepsilon \). Thus, \(\, x \in \overline {A} \, \implies \forall \, \varepsilon > 0, \, \big (x - \varepsilon \, ,\, x + \varepsilon \big ) \cap A \neq \emptyset \). Alternatively, \(\, x\not \in \overline {A}\,\implies \, \exists \varepsilon > 0\,\) for which \(\, \big (x - \varepsilon \, , \, x + \varepsilon \big ) \cap A = \emptyset \,\) or \(\, \exists \, \varepsilon > 0\, \ni \, \big (x - \varepsilon \, ,\, x + \varepsilon \big ) \subset A^C = \mathbb {R} - A\). Hence the following
theorem.
Theorem 1.3.2. If \(A\) is any subset of \(\mathbb {R}\), then \(\, x\not \in \overline {A}\, \iff \, x\in \big (A^C\big )^0\).
Thus, for any subset \(A\) of \(\mathbb {R},\, \big (\overline {A}\big )^C = \big (A^C\big )^0 \, \implies \, \overline {A} = \Big [\big (A^C\big )^0\Big ]^C\).
Proof. Both sides say the same thing about neighbourhoods of \(x\).
\(x\notin \overline {A}\) means \(x\) is neither in \(A\) nor a limit point of \(A\), which holds exactly when some \(\varepsilon >0\) gives \(\left (x-\varepsilon ,\,x+\varepsilon \right )\cap A = \emptyset \), that is \[\left (x-\varepsilon ,\,x+\varepsilon \right )\subseteq A^{C}.\] That last statement is precisely the assertion that \(x\) is an interior point of \(A^{C}\), i.e. \(x\in \left (A^{C}\right )^{0}\). The implications reverse at every step, so the two are equivalent. □
Remark. Written without the point, the identity is \(\overline {A}^{\,C} = \left (A^{C}\right )^{0}\), and taking complements gives \(\left (\overline {A}\right ) = \left (\left (A^{C}\right )^{0}\right )^{C}\). Closure and interior are dual under complementation, which is why every theorem about one has a mirror image about the other, and why it is enough to prove half of them.
Definition 1.3.3. A point \(x\in \mathbb {R}\) is called an exterior point of a set \(A\) if \(\, \exists \,\) a nbd \(N_x\) , of \(\, x \, \ni N_x \cap A = \emptyset \). The set of all
exterior points of \(A\) is called the exterior of \(A\) and is denoted by \(ext(A)\).
Definition 1.3.4. A point \(x\in \mathbb {R}\) is called a boundary point of a set \(A\) if it is neither an interior point
nor exterior point of \(A\). The set of all boundary points of \(A\) is called the boundary of \(A\) is denoted
by \(b(A)\).
Note.
- (i).
- \(\mathbb {R} = A^0 \cup b(A) \cup ext(A)\)
- (ii).
- \(ext(A) = \big (A^C\big )^0\)
- (iii).
- \(b(A) = \big [ A^0 \cup ext(A) \big ]^C\)
Definition 1.3.5. Let \(A\) be a subset of \(\mathbb {R}\). Then \(A\) is said to be an open set if every point of \(A\) is an
interior point of \(A\).
i.e \(\, A\) is an open set if for each \(\, x \in A\, \exists \, \varepsilon > 0\, \ni \, \big (x - \varepsilon \, , \, x + \varepsilon \big ) \subset A\).
Note. A subset \(A\) of \(\mathbb {R}\) is open if it is a neighborhood of all its points i.e if \(A = A^0\).
Theorem 1.3.6. For a subset \(A\) of \(\mathbb {R}\),
- (i).
- \(A\) is open \(\, \iff \, A^C\) is closed.
- (ii).
- \(A\) is closed \(\, \iff \, C^C\) is open.
Proof.
- (i).
- Suppose \(A\) is open. Then \(\, \forall \, x \in A\, \exists \, \varepsilon > 0\, \ni \, \big (x - \varepsilon \, , \, x + \varepsilon \big ) \subset A\). This means that each \(\, x\in A\, ,\, x\not \in A^c\) and \(\, \exists \varepsilon > 0\, \ni \big (x- \varepsilon \, ,\, x + \varepsilon \big ) \cap A^C = \emptyset \). Thus, for each \(\, x\in A\, x\) is not a limit point
of \(A^C\). Therefore, if \(y\) is a limit point of \(A^C\), then \(y\in A^C\). Hence, \(A^C\) is closed.
conversely, suppose \(A^C\) is closed. Let \(\, A^C = \mathbb {R}\). Then \(\, A = \emptyset \), which is open. Let \(A^C \neq \mathbb {R}\). Then \(A\neq \emptyset \). Let \(\, x\in A\). Then \(\, x\not \in A^C\,\) and \(\, x\in \mathbb {R}\). But since \(A^C\) is closed \(\, \big (A^C\big )^0 \, \implies \, x\in \big (A^C\big )^0\,\) and \(\, x\in \mathbb {R}\, \implies \exists \, \varepsilon > 0\, \ni \big (x - \varepsilon \, , \, x + \varepsilon \big ) \cap A^c = \emptyset \, \implies \exists \, \varepsilon > 0 \, \ni \, \big ( x - \varepsilon \, ,\, x + \varepsilon \big )\subset A\). Therefore \(A\) is open. - (ii).
- Prove this in similar way an exercise
However, it should be understood that if a set is not open it need not to be closed. A set is
open or closed according as it satisfies the criterion of the definition of open or closed
sets.
□
Corollary 1.3.7. For every subset \(A\) of \(\mathbb {R}\), \(\, A^0\,\) is the largest open subset of \(A\).
Proof. Suppose there exist another open subset \(\, G\ni G\subset A\). Let \(x\in G\). Then since \(G\) is open it is a nbd of \(x\). Thus
\(\, x\in A^0 \, \implies \, G\subset A^0\). Hence the result.
□
Proof. Let \(\, G = \big \{G_i:\, \forall i, \, G_i\,\) is open \(\big \}\) be an arbitrary family of open sets. Let \(\, G = \displaystyle {\bigcup _{i\in I}G_i}\,\) and \(\, x \in G\). Then \(\, \exists \, G_i \in G\ni x \in G_i \subset \, \implies \, \exists \, \varepsilon > 0 \ni \, \big (x - \varepsilon \, ,\, x + \varepsilon \big ) \subset G_i \subset G\). Therefore, \(G\) is open.
□
Proof. Let \(\, \displaystyle {G = \bigcap ^{n}_{i = 1} G_i}\ni G_i\,\) is open \(\, \forall i = 1,2,\cdots ,\cdots ,n\). If \(G= \emptyset \), then it is open.
Suppose \(\, G \neq \emptyset \). Then \(\, x \in G \, \implies \, x \in G_i\, \forall i = 1,2,\cdots \cdots , n\). Since, \(G_i\) is open for each \(i = 1,2, \cdots \cdots , n,\, \exists \varepsilon _i > 0 \, \forall i = 1,2, \cdots \cdots , n.\, \ni \, \big ( x - \varepsilon _i\, , \, x + \varepsilon _i\big ) \subset G_i\, \forall i = 1, 2, \cdots \cdots , n\). Thus, \(\exists \, \varepsilon = \min \big \{ \varepsilon _1, \, \varepsilon _2,\cdots \cdots ,\, \varepsilon _n\big \} \ni \big (x - \varepsilon \, , \,x + \varepsilon \big ) \subset G\). Therefore \(G\) is open.
However the theorem does not hold for intersection of an arbitrary family of open sets. For
example we defined \(G_n = \big (x - 1/n\, , \, x + 1/n\big ),\, n\in \mathbb {N},\, x\in \mathbb {R}\). This means that \(\, \bigcap _{n\in \mathbb {N}} G_n = \big \{x\big \}\), which is not an open set.
□
Definition 1.3.10. A set \(A\) is said to be dense (everywhere dense) in \(\mathbb {R}\) if \(\, \overline {A} = \mathbb {R}\). If \(\, x \in A\, \implies \, x \in A'\,\) i.e \(\, A\subset A'\), then \(A\) is said
to be dense in itself.
Definition 1.3.11. A set is said to be nowhere dense in \(\mathbb {R}\) if \(\, \big (\overline {A}\big )^0 = \emptyset \).
On the other hand, if every subinterval of an interval \(I\) has a part which does not contain any
point of \(A\) than \(A\) is said to be nowhere dense in \(I\).
In Case no subinterval of \(I\) is free from point of \(A\), then the set \(A\) is said to be everywhere dense in
\(I\).
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