# Thread: Verification of a topology

1. ## Verification of a topology

I have to show that

$\displaystyle \mathcal{T} = \left\{ (a, b] : a<b, a,b \in \mathbb{R}\right\}$

is a topology. I know that this is not a topology, because it forms the basis for a topology on $\displaystyle \mathbb{R}$. But why is this the case? I am thinking of the case, for instance, $\displaystyle (1,3] \cup (5,7]$. I don't think this is in $\displaystyle \mathcal{T}$, or am I incorrect?

HTale.

2. Originally Posted by HTale
I have to show that

$\displaystyle \mathcal{T} = \left\{ (a, b] : a<b, a,b \in \mathbb{R}\right\}$

is a topology. I know that this is not a topology, because it forms the basis for a topology on $\displaystyle \mathbb{R}$. But why is this the case? I am thinking of the case, for instance, $\displaystyle (1,3] \cup (5,7]$. I don't think this is in $\displaystyle \mathcal{T}$, or am I incorrect?

HTale.
You are right. Also,
Since $\displaystyle \left\{ (0, n]: n \in \mathbb{N}\right\} \in \mathcal{T}$, but $\displaystyle (0, \infty) \notin \mathcal{T}$, $\displaystyle \mathcal{T}$ is not closed under arbitrary union.

3. Originally Posted by HTale
I have to show that

$\displaystyle \mathcal{T} = \left\{ (a, b] : a<b, a,b \in \mathbb{R}\right\}$

is a topology. I know that this is not a topology, because it forms the basis for a topology on $\displaystyle \mathbb{R}$. But why is this the case? I am thinking of the case, for instance, $\displaystyle (1,3] \cup (5,7]$. I don't think this is in $\displaystyle \mathcal{T}$, or am I incorrect?

HTale.
As chabmgph said, you're correct.

For your information, $\displaystyle \mathcal{T}=\{(-\infty,a) ~:~ a \in \mathbb{R}\}$ does constitute a topology :P

4. Originally Posted by Moo
As chabmgph said, you're correct.

For your information, $\displaystyle \mathcal{T}=\{(-\infty,a) ~:~ a \in \mathbb{R}\}$ does constitute a topology :P
Does it? How about finite intersections of elements of $\displaystyle \mathcal{T}$?

5. Originally Posted by Moo
As chabmgph said, you're correct.

For your information, $\displaystyle \mathcal{T}=\{(-\infty,a) ~:~ a \in \mathbb{R}\}$ does constitute a topology :P
Originally Posted by HTale
Does it? How about finite intersections of elements of $\displaystyle \mathcal{T}$?
If $\displaystyle O_1, ..., O_n \in \mathcal{T}=\{(-\infty,a) ~:~ a \in \mathbb{R}\}$, then $\displaystyle O_1 = (-\infty, a_1) \text{ and } O_2 = (-\infty, a_2), ... \text{ with } a_1, a_2, ... \in \mathbb{R}$
Then $\displaystyle \bigcap_{i=1}^{n} O_i = (-\infty, m)$ where $\displaystyle m=\min\left\{ a_1, a_2, ...\right\}$.

6. Originally Posted by chabmgph
If $\displaystyle O_1, ..., O_n \in \mathcal{T}=\{(-\infty,a) ~:~ a \in \mathbb{R}\}$, then $\displaystyle O_1 = (-\infty, a_1) \text{ and } O_2 = (-\infty, a_2), ... \text{ with } a_1, a_2, ... \in \mathbb{R}$
Then $\displaystyle \bigcap_{i=1}^{n} O_i = (-\infty, m)$ where $\displaystyle m=\min\left\{ a_1, a_2, ...\right\}$.
OK, and also we have the analagous statement for unions, where we take the maximum. But, how about the empty set? You will never be able to get the empty set on this topology. The whole of the real line, yes but $\displaystyle \emptyset \notin \mathcal{T}$

Also, can I form the whole of the real line if I changed that to $\displaystyle (-\infty, a]$?

7. Oops sorry ! It's $\displaystyle \mathcal{T}=\{\emptyset~,~\mathbb{R} ~,~ (-\infty~,~a) ~:~ a \in \mathbb{R}\}$

8. Originally Posted by HTale
I have to show that

$\displaystyle \mathcal{T} = \left\{ (a, b] : a<b, a,b \in \mathbb{R}\right\}$

is a topology. I know that this is not a topology, because it forms the basis for a topology on $\displaystyle \mathbb{R}$. But why is this the case? I am thinking of the case, for instance, $\displaystyle (1,3] \cup (5,7]$. I don't think this is in $\displaystyle \mathcal{T}$, or am I incorrect?
Coming back to the original question, a set $\displaystyle \mathcal B$ is said to be a base (or basis) for a topology if the collection of (arbitrary) unions of members of $\displaystyle \mathcal B$ is a topology. The necessary and sufficient condition for a collection $\displaystyle \mathcal B$ of subsets of a space to form a base for a topology is that the union of all the members of $\displaystyle \mathcal B$ should be the whole space and that the intersection of any finite number of sets in $\displaystyle \mathcal B$ should also be in $\displaystyle \mathcal B$.

In the case of the set $\displaystyle \mathcal T$ above, it's easy enough to see that any finite intersection of half-open intervals is again a half-open interval. So $\displaystyle \mathcal T$ is the base for a topology (the half-open interval topology).

9. ## Re: Topology

Check out Chapter 2 of Part V of Topology and the Language of Mathematics (available for free here: Bobo Strategy - Topology). It gives some examples which you might find useful.