Show that $\tau$ is strictly finer than the euclidean topology.

Question

Let $A=\lbrace \frac{1}{n} : n \in \Bbb{N} \rbrace$, $H=\lbrace (a,b) \rbrace \cup \lbrace (a,b) - A \rbrace$. Let $\tau$ be the topology on $\Bbb{R}$ generated by the basis $H$.

First, I want to ask about $\tau$. How can I find it? I know $H$ is a basis for $\tau$, so any element of $\tau$ written as a union of element of $H$. Does this mean that $\tau = \bigcup \lbrace \lbrace (a,b) \rbrace \cup \lbrace (a,b) - A \rbrace \rbrace$ ?

I have another question about $H$. I think $\lbrace (a,b) -A \rbrace \subset (a,b)$, so is $\lbrace (a,b) \rbrace \cup \lbrace (a,b) - A \rbrace= \lbrace (a,b) \rbrace$ ?

Edit:

Show that $\tau$ is strictly finer than the Euclidean topology.

Let $\tau '$ is the Euclidean topology in $\Bbb{R}$.We know that $B=\lbrace (a,b):a,b \in \Bbb{R} \rbrace$ is a basis of $\tau '$, so any open set in $\tau '$ written as union of elements of $B$, then and from the definition of $H$, it is obvious that $B\subset H$, hence $\tau ' \subset \tau$. This means that $\tau$ is finer than the Euclidean topology. We have that $C=(-1,1)\setminus A$ is open in $\tau$, but is not in the Euclidean topology, because $0$ is element of $C$ but not an interior point to $C$. This means that $\tau$ is finer than the Euclidean topology. 0

Find $A'$. (The limit points of $A$).

Let $x \in \mathbb{R}$.

• If $x<0$, there exists a neighborhood of $x$, which is $(x-1,0)$, such that $(x-1,0)\cap A=\emptyset$. Thus $x\notin A’$ for all $x<0$.
• If $x=0$, there exists a neighborhood of $0$, which is $V=(-1,1)\setminus A$, such that $V\cap A=\emptyset$. Thus $0\notin A’$.

• If $0<x<1$, there exists a neighborhood of $x$, which is $U=(0,1)\setminus A$, such that $U\cap (A\setminus \lbrace x \rbrace)=\emptyset$. Thus $x\notin A’$.

• If $x=1$, we have that $(\frac{3}{4},2)$ is a neighborhood of $1$, and $(\frac{3}{4},2)\cap (A\setminus \lbrace 1 \rbrace )=\emptyset$. Hence $1 \notin A’$.

• If $x>1$, there exists a neighborhood of $x$, which is $(\frac{x+1}{2},x+1)$, and $(\frac{x+1}{2},x+1) \cap A=\emptyset$. Then $x\notin A’$.

As a result, $A’=\emptyset$.

Is that true, please? Thanks.

Answer 1

No, $\bigcup \lbrace \lbrace (a,b) \rbrace \cup \lbrace (a,b) - A \rbrace \rbrace = \lbrace (a,b) \rbrace \cup \lbrace (a,b) - A \rbrace$

{(a,b) - A} is not a subset of (a,b).
(a,b) - A is a subset of (a,b).
Also $\lbrace (a,b) \rbrace \cup \lbrace (a,b) - A \rbrace$ = { (a,b), (a,b)-A }

The topology is all sets of the form
$\cup${ (a,b) : (a,b) in S} $\cup$ $\cup${ (a,b) - A : (a,b) in S }
where S and T are subsets of R×R.

Answer 2

You don't have to "find" $\tau$, i.e. completely characterize all the $\tau$-open sets. All you have to do is exhibit just one set that is open w.r.t. $\tau$ but isn't open w.r.t. to the standard Euclidean topology. For this instance it might be easier to exhibit a $\tau$-closed set that isn't Euclidean-closed, which is equivalent.