Answer to Question #108197 in Real Analysis for priya

Question

a few questions:
1.Give a neighbourhood of a point 7.3
2. Is [0,1] an open set?

Answer 1

We will use the following definition of a base of topology: every nonempty open set is an union of elements from the base.

If we consider the standard topology on $\mathbb R$ (that is topology, base $\mathcal B$ of which is the set of open intervals $\{(a,b)|a<b\}$, then

1)$\bigcup\limits_{z\in\mathbb Z} \left(z,z+\frac{1}{2}\right)$ is a neighbourhood of $7.3$, because $7.3\in(7,7.5)\subset\bigcup\limits_{z\in\mathbb Z} \left(z,z+\frac{1}{2}\right)$ and $\bigcup\limits_{z\in\mathbb Z} \left(z,z+\frac{1}{2}\right)$ union of open intervals.

$(7,7.5)$ is also a neighbourhood of $7.3$, because $7.3\in(7,7.5)$ and $(7,7.5)$ is an open interval.

2)Prove that if a base $\mathcal B$ of topology is given, then equivalent definition of open sets is the following:

Nonempy $A$ is open set if and only if for every $x\in A$ there is $U_x\in\mathcal B$ such that $x\in U_x$ and $U_x\subset A$.

Let for every $x\in A$ there is $U_x\in\mathcal B$ such that $x\in U_x$ ​ and $U_x\subset A$ . Then we have $A=\bigcup\limits_{x\in A}\{x\}\subset\bigcup\limits_{x\in A}U_x$ and, since $U_x\subset A$ for every $x\in A$, we have $\bigcup\limits_{x\in A}U_x\subset A$. So $A=\bigcup\limits_{x\in A}U_x$, that is $A$ is open set by definition of a base of topology.

Now let $A$ be an nonempty open set. Take arbitrary $x\in A$. By the definition of a base of topology we have $A=\bigcup\limits_{Y\in\mathcal K}Y$ for some $\mathcal K\subset\mathcal B$. Since $x\in A$, there is $Y_0\in K$ such that $x\in Y_0$. Since $\mathcal K\subset\mathcal B$, we have $Y_0\in\mathcal B$. So $x\in Y_0$, where $Y_0\in B$ and $Y_0\subset A$.

Hence for the standard topology on $\mathbb R$ we have $A$ is an nonempty open set if and only if for every $x\in A$ there is an open inerval $(c,d)$ such that $x\in (c,d)$ and $(c,d)\subset A$.

Prove that $[0,1]$ is not an open set. Indeed, $0\in[0,1]$, but for every open interval $(c,d)$, which contains , we have $\frac{c}{2}\in(c,d)\setminus [0,1]$, that is $(c,d)\not\subset [0,1]$. So $[0,1]$ is not an open set.

Answer: 1)$\bigcup\limits_{z\in\mathbb Z} \left(z,z+\frac{1}{2}\right)$, $(7,7.5)$ are neighbourhoods of $7.3$ in stanard topology on $\mathbb R$

2)$[0,1]$ is not an open set in stanard topology on $\mathbb R$.