I came across this term while studying for a test and got curious as to what this actually is.
could someone please the concept in a relatively easy language?
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$\begingroup$As in the comments, clopen means both open and closed.
As an example, take a disconnected space $X$ like the union of two disjoint intervals, with the subspace topology, in $\Bbb R$. Then each interval would be clopen in $X$.
Clopenness is a topological notion. Remember, open sets are designated subsets of the space. Closed sets are then sets whose complements are open. Note that the two notions are not exclusive of each other.
$\endgroup$ $\begingroup$A "clopen" set meets both the definition of a closed set and an open set.
The classic examples are the real numbers and the empty set.
$\endgroup$ $\begingroup$I will give you my understanding within the narrow focus of considering all elements in the complex plane (i.e. $\mathbb{C}$).
This understanding is based exclusively on chapter 2 (The Rudiments of Plane Topology) from "An Introduction to Complex Function Theory" : 1992 : Bruce Palka.
$\mathbb{C}$ consists of all #'s of the form $a + ib$,
where $a$ and $b$ are real #'s and $i$ represents $\sqrt{-1}.$
A subset $S$ of $\mathbb{C}$ is deemed open, if and only if
every element in $S$ is an interior point of $S$.
An element $s \in S$ is deemed an interior point, if and only if
there exists some $\delta > 0$, such that
for all $z \in \mathbb{C},$ where $0 \leq |z - s| < \delta,$$z \in S.$
Here, given two complex #'s $z = x + iy$ and $w = u + iv,$
then |z - w| is defined to be $\sqrt{(x - u)^2 + (y - v)^2}.$
Given any subset $S$ of $\mathbb{C}$, its complement is defined to be
the set of all elements that are in $\mathbb{C}$ but are not in $S.$
A subset of $\mathbb{C}$ is deemed closed if and only if
its complement is deemed open.
Here, it is immediate that $\mathbb{C}$ and its complement $\phi$ (i.e. the empty set) are both open.
Consequently, both of these sets are also closed.
What remains is to prove that no other subsets of $\mathbb{C}$ can be simultaneously open and closed. To do this, I must introduce the notion of a boundary point of a set.
Given a subset $S$, a point $b$ (which may or may not be an element in $S$) is deemed a boundary point of $S$ if and only if
For any positive $\delta$, no matter how small, the set
{$z ~: ~0 \leq |z - b| < \delta$} will contain
at least one element that is in $S,$and at least one element that is not in $S.$
Two of the results from chapter 2 (The Rudiments of Plane Topology) are
(1)
If a set is closed then it must contain all of its boundary points.
(2)
If a set is open then it can not contain any of its boundary points.
Based on these two results, the only way a subset of $\mathbb{C}$ can simultaneously be open and closed is if the subset contains no boundary points.
Conjecture
To the best of my knowledge, the only such subsets of $\mathbb{C}$ are the set $\mathbb{C}$ itself and its complement $\phi$.
The following analysis represents a somewhat informal proof of the conjecture.
Let $S$ represent a proper subset of $\mathbb{C}$, and let $T$ represent the complement of $S.$ This means that neither $S$ nor $T$ are empty sets.
Let $s$ be any element in $S$, and let $t$ be any element in $T.$
Let $L$ denote all elements on the line segment that connect $s$ and $t$.
That is, $L ~\equiv~${$ z ~: ~z = s + [a\times (t-s)] ~: ~a \in \mathbb{R}, ~0 \leq a \leq 1$}.
Clearly, when examining the elements of $L$ of the form $s + [a\times (t-s)],$
when $a = 0$, the element equals $s$, which is in $S,$and when $a = 1$, the element equals $t$, which is in $T.$
Consider the set $B$ of all of the scalars $b$ such that
$0 \leq b \leq 1$ and
all elements in $L$ of the form $s + [b\times (t-s)],$ are in $S$.
Let $c$ denote the smallest element in $B$
such that for any $\delta > 0$,
the interval $(c - \delta, c]$ contains
at least one element that is in $B$ and
the interval $[c, c+\delta)$ contains
at least one element that is not in $B.$
Then the element in $L$ represented by $[s + c\times (t-s)],$ must bea boundary point of $S.$
The above analysis hinges on proving that regardless of the nature of the underlying set $S$, a subset of $\mathbb{C},$ the scalar $c$ must exist. My knowledge of topology is way too unsophisticated to discuss proving the existence of the scalar $c.$
A minor variation of the above analyis is that since $\mathbb{R}$ is deemed a complete ordered field, and since $B$ is a bounded subset of $\mathbb{R}$, there is a real # $d$ that is the least upper bound for $B.$Under this presumption, then $[s + d\times (t-s)]$ is (also) a boundary point for $S.$
Again, I lack the knowledge or sophistication to discuss the above presumption.
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