# Limit Points and Closure in a Topological Space ... Singh, Theorem 1..3.7 ... ...

#### Peter

##### Well-known member
MHB Site Helper
I am reading Tej Bahadur Singh: Elements of Topology, CRC Press, 2013 ... ... and am currently focused on Chapter 1, Section 1.2: Topological Spaces ...

I need help in order to fully understand Singh's proof of Theorem 1.3.7 ... (using only the definitions and results Singh has established to date ... see below ... )

In the above proof by Singh we read the following:

" ... ... So $$\displaystyle U$$ is contained in the complement of $$\displaystyle A \cup A'$$, and hence $$\displaystyle A \cup A'$$ is closed. It follows that $$\displaystyle \overline{A} \subseteq A \cup A'$$ ... ... "

My question is as follows:

Why does $$\displaystyle U$$ being contained in the complement of $$\displaystyle A \cup A'$$ imply that $$\displaystyle A \cup A'$$ is closed ... and further why then does it follow that $$\displaystyle \overline{A} \subseteq A \cup A'$$ ... ...

Help will be appreciated ...

Peter

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It is important that any proof of the above remarks only rely on the definitions and results Singh has established to date ... namely Definition 1.3.3, Proposition 1.3.4, Theorem 1.3.5 and Definition 1.3.6 ... which read as follows ... :

Hope that helps ...

Peter

#### Opalg

##### MHB Oldtimer
Staff member
Why does $$\displaystyle U$$ being contained in the complement of $$\displaystyle A \cup A'$$ imply that $$\displaystyle A \cup A'$$ is closed ...
The key to this lies in Theorem 1.3.5, which says that $x\in\overline{A} \Longleftrightarrow U\cap A\ne\emptyset$ for every (open) nbd of $x$. [Note that the last symbol in that statement is $x$, not $X$ as Singh's book mistakenly has it.]

Replace $A$ by $A\cup A'$ in the statement of 1.3.5 to see that $x\in\overline{A\cup A'} \Longleftrightarrow U\cap (A\cup A')\ne\emptyset$ for every (open) nbd of $x$. But the proof of 1.3.7 shows that if $x\notin A\cup A'$ then $x$ has an open nbd $U$ contained in the complement of $$\displaystyle A \cup A'$$, so that $U\cap (A\cup A') = \emptyset$. That contradicts that criterion in 1.3.5 for $x$ to be in $\overline{A\cup A'}$. In other words, $x\notin A\cup A' \Rightarrow x\notin \overline{A\cup A'}$, the contrapositive of which is $x\in \overline{A\cup A'} \Rightarrow x\in A\cup A'$. Therefore $\overline{A\cup A'} \subseteq A\cup A'$. The reverse inclusion is obvious, and so $\overline{A\cup A'} = A\cup A'$, meaning that $A\cup A'$ is closed.

... and further why then does it follow that $$\displaystyle \overline{A} \subseteq A \cup A'$$ ...
$A\cup A'$ is a closed set containing $A$. It is therefore larger than the smallest such set, which by definition is $\overline{A}$.

#### Peter

##### Well-known member
MHB Site Helper
The key to this lies in Theorem 1.3.5, which says that $x\in\overline{A} \Longleftrightarrow U\cap A\ne\emptyset$ for every (open) nbd of $x$. [Note that the last symbol in that statement is $x$, not $X$ as Singh's book mistakenly has it.]

Replace $A$ by $A\cup A'$ in the statement of 1.3.5 to see that $x\in\overline{A\cup A'} \Longleftrightarrow U\cap (A\cup A')\ne\emptyset$ for every (open) nbd of $x$. But the proof of 1.3.7 shows that if $x\notin A\cup A'$ then $x$ has an open nbd $U$ contained in the complement of $$\displaystyle A \cup A'$$, so that $U\cap (A\cup A') = \emptyset$. That contradicts that criterion in 1.3.5 for $x$ to be in $\overline{A\cup A'}$. In other words, $x\notin A\cup A' \Rightarrow x\notin \overline{A\cup A'}$, the contrapositive of which is $x\in \overline{A\cup A'} \Rightarrow x\in A\cup A'$. Therefore $\overline{A\cup A'} \subseteq A\cup A'$. The reverse inclusion is obvious, and so $\overline{A\cup A'} = A\cup A'$, meaning that $A\cup A'$ is closed.

$A\cup A'$ is a closed set containing $A$. It is therefore larger than the smallest such set, which by definition is $\overline{A}$.

Thanks for the help, Opalg ... really appreciate it ...

Just working through what you have written ...

Thanks again ...

Peter