What does it mean for a set to be open in a Topology? "What does it mean for a set to be open in a Topology?" - I have a hard time getting my head around what it means.
Example: 
Is the set $(0,1]\subset\mathbb{R}$ open in the standard Topology on $\mathbb{R}$ or open in the Topology generated by the basis $B = \{[a,b)\subset \mathbb{R} | a < B \}$ on $\mathbb{R}$?
My answer / idea:
I would think that since the standard topology is generated by open intervals, we could have the interval $(0,2)$, which contains $(0,1]$ - so I guess that it would be "open" in the topology.
For the Topology on the basis B, I would also say that since $(0,1] \subset [0,2)$ it must also be open in that topology.
Though, I have a clear feeling that I have the wrong idea. I know that the interval being simply contained in the Topology, is not an answer.
 A: Recall the definition. A topological space is a pair $(X,\mathcal{T})$where $X$ is a set and $\mathcal{T}$ (is defined to be the set of open sets,) $\mathcal{T}$ is a collection of subsets of $X$ such that


*

*$\emptyset\in\mathcal{T}$

*Any union $A$ of a collection of $A_i\in\mathcal{T}$ is also in $\mathcal{T}$

*Any finite intersection $A$ of $A_1,\ldots,A_n\in\mathcal{T}$ is also in $\mathcal{T}$.


Then the topology generated by a set $\mathcal{A}$ is the coarsest ("smallest") topology $\mathcal{T}$ such that $\mathcal{A}\subseteq\mathcal{T}$.
So an open space in the topology genrated by the intervals $B=\{[a,b)\subseteq\Bbb{R}:a<b\}$ has the open sets $A\in\mathcal{T}$ such that $A$ can be obtained via taking arbitrary unions or finite intersections of $B$-intervals. For example, $[0,2)$ is open, as is $\bigcup_{n\ge 1}[0,n)=[0,\infty)$. For a more interesting example, consider the set given by
$$\bigcup_{n\ge 2}\,[1/n,1),$$
which is also open, as an arbitrary union of open sets; notice this union is equal to $(0,1)$.

On a side note, $(0,1]$ is in fact not open. To show a set is not open, one way is to show that the complement is not closed. In your example, to show that $(0,1]$ is not open in the topology generated by $B$, I would show that $(-\infty,0]\cup(1,\infty)$ is not closed in the topology generated by $B$. 
Lemma. A subset $C$ is closed iff every convergent (in $X$) sequence $x_n\to x$ with points $x_n\in C$ has $x\in C$. 
To show that $x_n\to x$ in the topology generated by $B$ it is sufficient to show that for every set in the basis $[a,b)\in B$, that contains $1$, there is some $N$ such that $n\ge N$ implies $x_n\in [a,b)$. Let $x_n=1+1/n$, I claim that $x_n\to 1$ in $(\Bbb{R},B)$. For every base element containing $1$ (i.e. $[1,1+\varepsilon)$), there is an $N$ such that $n\ge N$ implies $1/n<\varepsilon$, i.e. $x_n\in [1,1+\varepsilon)$; but $1\notin(-\infty,0]\cup(1,\infty)$. Hence, $(-\infty,0]\cup(1,\infty)$ is not closed, so $(0,1]$ is not open.
