is a common shorthand for "the set of all subsets of X". That is the definition of "discrete topology".
Are power sets always a discrete topology?
One video lecture I watched introduced a topology as for some set .
(I know this forum is more for discussion not help but this is extracurricular work so I figured no need to take up space in the forum where people have questions for assignments)
A topology is a certain KIND of subset of $2^{X}$, for a set $X$.
One possible topology for $X$ is $2^{X}$, this is known as the discrete topology on $X$.
Another possible topology is $\{\emptyset,X\}$, this is called the indiscrete topology on $X$.
Often, a topology is somewhere in-between.
For example, if we have a metric, or DISTANCE function, $d: X\times X \to \Bbb R_0^+$ we can form the subsets of $X$, that have this property:
$U \in \mathfrak{T} \iff \forall u \in U,\ \exists B_{\epsilon}(u) \subseteq U$, where $B_{\epsilon}(u) = \{x \in X: d(u,x) < \epsilon\}$
The $B_{\epsilon}(u)$ is called "an epsilon-ball centered at $u$", and in the case:
$X = \Bbb R^3$, and $d((x_1,y_1,z_1),(x_2,y_2,z_2)) = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2 - z_1)^2}$ is literally "an open ball of radius $\epsilon$"
so an open set $U$ in this topology is one that we can "squeeze in an open epsilon-ball" in, at any point (we may have to choose $\epsilon$ fairly small).
In the case where $X = \Bbb R$ ,and $d(x_1,x_2) = |x_2 - x_1|$, such an "epsilon ball" is just the open interval $(u-\epsilon,u+\epsilon)$, and yield the same topology as that generated by unions of open intervals.
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Often, what we are interested in, is points we consider "near" a point $x \in X$. A distance function is one way of being precise about this.
In the absence of a metric function, we use neighborhoods of points to define nearness. Basically, neighborhoods of a point $x$ act as a way of "zeroing in" on $x$, by taking "smaller neighborhoods". Since we may not have a metric to "give us a number", we use "containment" as a way of comparing "closeness". This only works well if we have "the right amount of neighborhoods".
For example, if we have the discrete topology, all the points are "not near" each other. The example you should think of here, is the set of integers, where $k$ and $k+1$ are "very separated".
By contrast, if we have the indiscrete topology, everything gets rolled up into "one neighborhood", which means we have no way of telling points apart. It is sort of like seeing 3-space as a solid chunk of ice, that we cannot penetrate into to measure anything at all.
There are often many possible topologies on any given "space", some of them are rather bizarre. Here is one that is "halfway bizarre", it's a bit unusual, but used often to test theorems.
In it, we take $X = \Bbb R$, the real numbers. We say $U \in\mathfrak{T}$ if $T -U$ is finite, or $U = \emptyset$ or $U = \Bbb R$. This is called the co-finite, or finite complement, topology. In this topology, the only closed sets are finite sets (but not all infinite sets are open).
okay, very interesting. My book has only touched on discrete topologies and indiscrete topologies so far. I was curious about a topology that was perhaps neither discrete, nor indiscrete (if thats correct to say).
Can a topology be partitioned similarly to how a set can be partitioned?
Kind of. Recall that a partition of a set induces an equivalence relation. Similarly, an equivalence relation on a set induces a partition.
Now a topological space is TWO things: a set $X$, and a topology $\mathfrak{T}$ on $X$.
We can turn a partitioned set $X/\sim$ into a topological space "based on $X,\mathfrak{T}$" in the following way:
We take the open sets in $X/\sim$ to be the sets whose pre-images under the mapping $q:x \to [x]$ are open in $X$.
This is sometimes hard to wrap one's head around, so here is an example:
Suppose we have the mapping $f:\Bbb R \to \Bbb R$ given by $f(x) = x^2$. We'll use "the usual topology" generated by open intervals on $\Bbb R$.
We can partition $\Bbb R$ by saying $x \sim y$ if $x^2 = y^2$. So if $x \neq 0$, then $[x] = \{x,-x\}$, and $[0] = \{0\}$.
Topologically speaking, we've folded the real line in half, bending it over at the origin.
So let's say $A$ is a subset of $\Bbb R/\sim$. It's pre-image, under $f$ is a set $B \cup -B$ where $-B = \{x \in \Bbb R: -x\in B\}$. For this to be open, $B$ has to be open.
We can think of $\Bbb R/~$ as "essentially" $\Bbb R_0^+$ since the mapping $[x] \mapsto x$ is a bijection when $x \geq 0$.
So the open sets in $\Bbb R/~$ correspond to unions of open intervals $(a,b)$ where $0 \leq a \leq b$. Their pre-images are unions of pairs of open intervals: $(-b,-a) \cup (a,b)$.
We've "collapsed" matching pairs of open intervals in the topology for $\Bbb R/\sim$ the same way we "collapsed" $\{-x,x\}$ into $[x]$.
Note that for any $x$, we have $f$ constant on $[x]$.
This same idea, is used quite frequently in topology. For example, we can take the real line, and "wrap it around a circle", identifying all points that are $2\pi$ apart.
The "basic" open sets on the circle, are then "open arcs", which have pre-images of equal sized open intervals, spaced $2\pi$ apart across the entire real line.