Show that the half open interval is a topology

[DotKite]

Homework Statement

We are given $H$ = {$O | \forall x, \exists a,b \in R$ s.t $x \in [a,b] \subseteqq O$}$\bigcup {\oslash}$

and are asked to show that it is a topology on R

Homework Equations

Definition of a topological space

The Attempt at a Solution

I am trying to convince myself that R is in H. I do not see it. This is, of course, the first property of a topological space.

 

[R136a1]

So take $x\in \mathbb{R}$, you must find $a,b$ such that $x\in [a,b]\subseteq \mathbb{R}$. Any idea of what the $a$ and $b$ are going to be?? (Hint: any $a$ and $b$ are good as long as $a\leq b$)

 

[Dick]

Homework Statement

We are given $H$ = {$O | \forall x, \exists a,b \in R$ s.t $x \in [a,b] \subseteqq O$}$\bigcup {\oslash}$

and are asked to show that it is a topology on R

Homework Equations

Definition of a topological space

The Attempt at a Solution

I am trying to convince myself that R is in H. I do not see it. This is, of course, the first property of a topological space.

Since for any x, x is an element of [x,x], the way you stated the first part of your union, the only subset O of R that will satisfy that is R itself. There is something wrong with your problem statement. That doesn't look like a half-open topology to me.

 

[DotKite]

OK so now I am trying to show that a finite intersection is in H. I am still really confused on what this set actually is. I would appreciate if anyone could explain to me how this set is/works.

 

[vela]

Homework Statement

We are given $H$ = {$O | \forall x, \exists a,b \in R$ s.t $x \in [a,b] \subseteqq O$}$\bigcup {\oslash}$

Why does the thread title refer to the half-open interval topology? I don't see any half-open intervals in your definition.

You do realize that for any set A, its union with the empty set is itself, right? So you might as well have written
$$H = \{ O: \forall x, \exists a,b \in \mathbb{R} \text{ such that } x \in [a,b] \subset O \}.$$ What you probably meant was
$$H = \{ O: \forall x, \exists a,b \in \mathbb{R} \text{ such that } x \in [a,b] \subset O \} \cup \{\emptyset\},$$ so that the empty set is an element of H.

Finally, what about the point Dick brought up? The only set that satisfies your definition is $\mathbb{R}$ since $x$ can be anything, so this topology is the trivial topology $\{ \emptyset, \mathbb{R}\}$, not the half-open interval topology.

You really need to get the definition of H down before you go about trying to prove anything with it.

 

[Dick]

The half-open interval topology or lower limit topology is usually defined as the topology generated by the basis sets [a,b). Your set definition doesn't resemble that in the slightest.

 

[DotKite]

The half-open interval topology or lower limit topology is usually defined as the topology generated by the basis sets [a,b). Your set definition doesn't resemble that in the slightest.

My mistake people. There was suppose to be an open bracket at b. [a,b)

 

[Dick]

My mistake people. There was suppose to be an open bracket at b. [a,b)

Ok, so your topology consists of the set of all sets O such for every point x in O, there is an interval [a,b) that contains the point x and is a subset of O. Does that make it clearer what your topology is?