Okay so $\mathbb{R}$ is a sequential space, so limit points of sequences coincide with topological limit points (I think......someone blow your horns at me if this is wrong), the sequential limit point of $A$ if $\{0\}$ so $A' = \{0\}$, and hence $\bar{A} = A \cup \{0\}$
@orbit-stabilizer A set in $\Bbb R$ is open if, for every point in the set, there exists an open interval surrounding the point contained in the set. Right?
I think that's all of them. Anything that satisfies those rules is called a topology; $X$ is the base space, and $\tau$ is the set of open sets in $X$.
@AkivaWeinberger Yeah that's all, just could be confusing to say "closed" under finite intersection etc. when there's another topological meaning for closed
Thus, we can define continuous functions between any two topological spaces $(X,\tau)$ and $(Y,\upsilon)$ by saying a function $f:X\to Y$ is continuous iff for any $U\in\upsilon$, $f^{-1}(U)$ is in $\tau$
@orbit-stabilizer To define the order topology, you define $(a,b)$ to be $\{x:a<x<b\}$, and define $(-\infty,b)$ and $(a,\infty)$ similarly, and then call those "open intervals",
@orbit-stabilizer Let $(X,\tau)$ be a topology. Let $G$ be a group that acts on $X$. Then, define the equivalence relation $x \sim y \equiv ([x]=[y])$ where $[x]$ denotes the orbit of $x$. Then, the quotient space is $X/\sim$
In $[0,1]$, if we have a sequence $(a_0,a_1,a_2,\dots)$ such that all the $a_i$ are in $(0,1]$, we don't know where the limit will end up; it could be in $(0,1]$ and it could be in $\{0\}$.
