Neighbourhood, Open Sets, Closed Sets
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Exploring Open Sets in Real Analysis
Welcome to our comprehensive guide on Real Open Sets, a cornerstone concept in real analysis. In this article, you will discover precise definitions, illustrative examples, and essential theorems that elucidate the properties of open sets. We employ standard mathematical notation such as \(\mathbb{R}\) and \(\operatorname{N}_{\delta}(x)\) to ensure clarity and precision, while our engaging explanations are designed to cater to a broad audience.
Open Set
Definition: [Open Set]
Let \(S\subseteq \mathbb{R}\). \(S\) is said to be an open set if all the points in \(S\) is an interior point of \(S\).
That is \[S\subseteq\operatorname{Int} S. \]
Examples
1. The entire set \( \mathbb{R} \) is an open set because every real number is surrounded by an open interval contained in \( \mathbb{R} \).
2. The set of rational numbers \( \mathbb{Q} \) is not open; it fails to provide an open interval entirely within \( \mathbb{Q} \) around each of its points.
3. The natural numbers \( \mathbb{N} \) do not form an open set since each natural number lacks a surrounding open interval that lies solely within \( \mathbb{N} \).
4. Similarly, the set of integers \( \mathbb{Z} \) is not an open set.
5. Any finite set is not open because no point can have an open neighbourhood that remains completely inside the set.
6. Every open interval, such as \( (a, b) \), is an open set.
7. A closed interval, for instance \( [a, b] \), is not open since the endpoints lack open neighbourhoods contained in the set.
8. The null set \( \varnothing \) is considered open by definition.
Properties
1. \( \operatorname{Int} S =S \).
Theorems Concerning Open Sets
Theorem-1
Statement:
The union of two open sets is an open set.
Proof:
Let \(A\) and \(B\) be two open sets.
To prove \(A \cup B\) is open.
Let \[x \in A \cup B \implies x \in A \lor x \in B\].
Implies \(x\) is an interior point of \(A\) or \(B\) since \(A\) and \(B\) are open.
Then there exists \(\delta_{1},\delta_{2} \gt 0 \) such that
\[ \operatorname{N}_{\delta_{1}}(x) \subseteq A \lor \operatorname{N}_{\delta_{2}}(x) \subseteq B\]
Let \(0\lt \delta \lt \operatorname{min} \set{ \delta_{1},\delta_{2} } \) then
\[ \begin{align} & \operatorname{N}_{\delta}(x) \subseteq \operatorname{N}_{\delta_{1}}(x) \subseteq A \lor \operatorname{N}_{\delta}(x) \subseteq \operatorname{N}_{\delta_{2}}(x) \subseteq B \nonumber \\ \implies & \operatorname{N}_{\delta}(x) \subseteq A \cup B \nonumber \end{align}\]
Therefore \(x\) is an interior point of \(A \cup B \) then \[A \cup B \subseteq\operatorname{Int} (A \cup B). \]
Therefore \(A \cup B\) is open.
Theorem-2
Statement:
The intersection of two open sets is an open set.
Proof:
Let \(A\) and \(B\) be two open sets.
To prove \(A \cap B\) is open.
Let \[x \in A \cap B \implies x \in A \land x \in B\].
Implies \(x\) is an interior point of \(A\) and \(B\) since \(A\) and \(B\) are open.
Then there exists \(\delta_{1},\delta_{2} \gt 0 \) such that
\[ \operatorname{N}_{\delta_{1}}(x) \subseteq A \land \operatorname{N}_{\delta_{2}}(x) \subseteq B\]
Let \(0\lt \delta \lt \operatorname{min} \set{ \delta_{1},\delta_{2} } \) then
\[ \begin{align} & \operatorname{N}_{\delta}(x) \subseteq \operatorname{N}_{\delta_{1}}(x) \subseteq A \land \operatorname{N}_{\delta}(x) \subseteq \operatorname{N}_{\delta_{2}}(x) \subseteq B \nonumber \\ \implies & \operatorname{N}_{\delta}(x) \subseteq A \cap B \nonumber \end{align}\]
Therefore \(x\) is an interior point of \(A \cap B \) then \[A \cap B \subseteq\operatorname{Int} (A \cap B). \]
Therefore \(A \cap B\) is open.
Theorem-3
Statement:
The union of a finite collection of open sets is an open set.
Proof:
Let \(\set{A_{i}: i=1,2,..n}\) be a finite collection of open sets.
To prove \( \displaystyle\bigcup_{i=1}^{n} A_{i} \) is open.
Let \[x \in \displaystyle\bigcup_{i=1}^{n} A_{i}.\]
Then there exists atleast one \(p\in \mathbb{N}\) such that \[x\in A_{p},~1\leq p \leq n .\]
Implies \(x\) is an interior point of \(A_{p} \) since \(A_{p} \) is open.
Then there exists \(\delta \gt 0 \) such that
\[ \begin{align} & \operatorname{N}_{\delta}(x) \subseteq A_{p} \subseteq \displaystyle\bigcup_{i=1}^{n} A_{i}\nonumber \end{align}\]
Therefore \(x\) is an interior point of \(\displaystyle\bigcup_{i=1}^{n} A_{i} \) then \[\displaystyle\bigcup_{i=1}^{n} A_{i} \subseteq\operatorname{Int} \displaystyle\bigcup_{i=1}^{n} A_{i}. \]
Therefore \(\displaystyle\bigcup_{i=1}^{n} A_{i}\) is open.
Theorem-4
Statement:
The intersection of a finite collection of open sets is an open set.
Proof:
Let \(\set{A_{i}: i=1,2,..n}\) be a finite collection of open sets.
To prove \( \displaystyle\bigcap_{i=1}^{n} A_{i} \) is open.
Let \[x \in \displaystyle\bigcap_{i=1}^{n} A_{i}.\]
Then \[x\in A_{p}~\forall~ p=1,2,…,n .\]
Implies \(x\) is an interior point of \(A_{p}~\forall~ p=1,2,…,n \) since \(A_{p}~\forall~ p=1,2,…,n \) are open.
Then there exists \(\delta_{p} \gt 0,~ p=1,2,…,n \) such that
\[ \begin{align} & \operatorname{N}_{\delta_{p}}(x) \subseteq A_{p}~\forall~ p=1,2,…,n \nonumber \end{align}\]
Let \(0\lt \delta \lt \operatorname{min}\set{\delta_{p},~ p=1,2,…,n } \)
\[ \begin{align} & \operatorname{N}_{\delta}(x) \subseteq \displaystyle\bigcap_{i=1}^{n} A_{i}\nonumber \end{align}\]
Therefore \(x\) is an interior point of \(\displaystyle\bigcap_{i=1}^{n} A_{i} \) then \[\displaystyle\bigcap_{i=1}^{n} A_{i} \subseteq\operatorname{Int} \displaystyle\bigcap_{i=1}^{n} A_{i}. \]
Therefore \(\displaystyle\bigcap_{i=1}^{n} A_{i}\) is open.
Theorem-5
Statement:
The union of an arbitrary collection of open sets is an open set.
Proof:
Let \(\set{A_{i}: i\in I}\) be an arbitrary collection of open sets where \(I \) be an Index set.
To prove \( \displaystyle\bigcup_{i\in I}^{} A_{i} \) is open.
Let \[x \in \displaystyle\bigcup_{i\in I}^{} A_{i}.\]
Then there exists atleast one \( p\in I\) such that \[x\in A_{p}.\]
Implies \(x\) is an interior point of \(A_{p} \) since \(A_{p} \) is open.
Then there exists \(\delta \gt 0 \) such that
\[ \operatorname{N}_{\delta}(x) \subseteq A_{p} \subseteq \displaystyle\bigcup_{i\in I}^{} A_{i} \]
Therefore \(x\) is an interior point of \(\displaystyle\bigcup_{i\in I}^{} A_{i} \) then \[\displaystyle\bigcap_{i\in I}^{} A_{i} \subseteq\operatorname{Int} \displaystyle\bigcup_{i\in I}^{} A_{i}. \]
Therefore \(\displaystyle\bigcap_{i\in I}^{} A_{i}\) is open.
Important Note
1. The intersection of an arbitrary collection of open sets is not always open.
Consider the collection \(A_{n}=\left( -\frac{1}{n},\frac{1}{n} \right),~n\in \mathbb{N} \) of open sets.
Then \(\displaystyle\bigcap_{i\in \mathbb{N}}^{} A_{i}=\set{0}\) is not an open set since \(A_{1}\supseteq A_{2}\supseteq \cdots\).
Consider the collection \(A_{n}=\left( -n,n \right),~n\in \mathbb{N} \) of open sets.
Then \(\displaystyle\bigcap_{i\in \mathbb{N}}^{} A_{i}=(-1,1)\) is an open set since \(A_{1}\subseteq A_{2}\subseteq \cdots\).
Conclusion
In conclusion, the study of Real Open Sets is fundamental to the field of real analysis. By defining open sets, examining concrete examples, and proving key theorems, we develop a deeper understanding of the underlying topology of \(\mathbb{R}\). This exploration not only strengthens the conceptual framework but also sets the stage for advanced topics such as continuity and convergence. Our discussion has been carefully crafted to optimize search engine visibility and ensure clarity for learners at all levels.
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