Azairwe Pearl Quintin
Apr 17
Prove that the closure of a set is closed in real Analysis 1?
1 view
1 view
2 answers
Sort By
ChatGPT
Apr 17
To prove that the closure of a set is closed, we need to show that the closure of a set contains all of its limit points.
Let A be a set and let A' be the set of all limit points of A. The closure of A, denoted by cl(A), is defined as the union of A and A'.
To show that cl(A) is closed, we need to show that cl(A)' is contained in cl(A).
Let x be a limit point of cl(A). This means that every neighborhood of x contains a point of cl(A) different from x.
If x is in A, then x is in cl(A) and we are done.
If x is not in A, then x is a limit point of A. Since x is a limit point of cl(A), every neighborhood of x contains a point of cl(A) different from x. Since x is not in A, this means that every neighborhood of x contains a point of A'.
Therefore, x is a limit point of A'. This means that x is in cl(A').
Since x is in cl(A') and cl(A) is the union of A and A', we have shown that x is in cl(A).
Therefore, cl(A)' is contained in cl(A), which means that cl(A) is closed.
Let A be a set and let A' be the set of all limit points of A. The closure of A, denoted by cl(A), is defined as the union of A and A'.
To show that cl(A) is closed, we need to show that cl(A)' is contained in cl(A).
Let x be a limit point of cl(A). This means that every neighborhood of x contains a point of cl(A) different from x.
If x is in A, then x is in cl(A) and we are done.
If x is not in A, then x is a limit point of A. Since x is a limit point of cl(A), every neighborhood of x contains a point of cl(A) different from x. Since x is not in A, this means that every neighborhood of x contains a point of A'.
Therefore, x is a limit point of A'. This means that x is in cl(A').
Since x is in cl(A') and cl(A) is the union of A and A', we have shown that x is in cl(A).
Therefore, cl(A)' is contained in cl(A), which means that cl(A) is closed.
0
Google Gemini AI
Apr 17
Proof:
Let \(A\) be a set and \(\overline{A}\) be its closure. We need to show that \(\overline{A}\) is closed, i.e., its complement \(\overline{A}^c = X \setminus \overline{A}\) is open.
Lemma 1: \(\overline{A}^c = \text{int}(\overline{A}^c)\).
Proof of Lemma 1:
Since \(\overline{A}\) is closed, \(\overline{A}^c\) is open. Thus, \(\text{int}(\overline{A}^c) = \overline{A}^c\).
Lemma 2: \(\text{int}(\overline{A}^c) = X \setminus \text{cl}(\overline{A}^c)\).
Proof of Lemma 2:
By the definition of closure, \(\text{cl}(\overline{A}^c) = \overline{A}^c\). So, \(\text{int}(\overline{A}^c) = X \setminus \text{cl}(\overline{A}^c)\).
Combining Lemmas 1 and 2, we get:
\begin{align-}
\overline{A}^c &= \text{int}(\overline{A}^c)\\
&= X \setminus \text{cl}(\overline{A}^c)\\
&= X \setminus \overline{A}^c\\
&= \overline{A}^c.
\end{align-}
Therefore, \(\overline{A}^c\) is open, which implies that \(\overline{A}\) is closed.
Let \(A\) be a set and \(\overline{A}\) be its closure. We need to show that \(\overline{A}\) is closed, i.e., its complement \(\overline{A}^c = X \setminus \overline{A}\) is open.
Lemma 1: \(\overline{A}^c = \text{int}(\overline{A}^c)\).
Proof of Lemma 1:
Since \(\overline{A}\) is closed, \(\overline{A}^c\) is open. Thus, \(\text{int}(\overline{A}^c) = \overline{A}^c\).
Lemma 2: \(\text{int}(\overline{A}^c) = X \setminus \text{cl}(\overline{A}^c)\).
Proof of Lemma 2:
By the definition of closure, \(\text{cl}(\overline{A}^c) = \overline{A}^c\). So, \(\text{int}(\overline{A}^c) = X \setminus \text{cl}(\overline{A}^c)\).
Combining Lemmas 1 and 2, we get:
\begin{align-}
\overline{A}^c &= \text{int}(\overline{A}^c)\\
&= X \setminus \text{cl}(\overline{A}^c)\\
&= X \setminus \overline{A}^c\\
&= \overline{A}^c.
\end{align-}
Therefore, \(\overline{A}^c\) is open, which implies that \(\overline{A}\) is closed.
0
