Is every subset of $X$ compact and connected under the trivial topology on $X$?

Question

I was thinking about the trivial topology, in which the only open sets are $X$ and $\emptyset$, where $X$ is the entire space we are working in.

My doubt is: are all the subset of $X$ compact and connected? I understand that they are all closed, since the only open sets are $X$ and $\emptyset$. But compact means also bounded, so if for example $X = \mathbb{R}$, how can be, for example, $(1, +\infty)$ bounded?

Also I'm having problems in understanding connectedness. Say we have $[1, 3]\cup [4, 6]$, how is this connected in the trivial topology?

Answer 1

"... compact also means bounded ..." is true in a metric space, where the concept of bounded is defined. But in a topological space where no metric is defined, boundedness is undefined. So you have to adjust your intuition here so as to entirely ignore the concept of boundedness in approaching this problem. Instead, you have to directly apply the definition of compactness.

Compactness, by definition, says every open cover has a finite subcover. Well, in the trivial topology there are only 2 open sets, namely the empty set and the whole space. So, any given open cover is already finite, because that open cover has at most 2 elements. The space is therefore compact.

Disconnectedness, by definition, says that there is a disjoint pair of nonempty open sets whose union is the whole space. Well, in the trivial topology there is only 1 nonempty open set, namely the whole space. So there does not even exist two nonempty open sets (let alone two which are disjoint and whose union is the whole space). So the space is not disconnected --- that is to say, it is connected.

Answer 2

In the trivial topology, where the only open sets are the entire space $X$ and the empty set $\emptyset$, let's address your questions:

$\textbf{Compactness:}$ Every subset of $X$ is compact in the trivial topology. Compactness in a topology is not equivalent to boundedness in the metric space sense. In the trivial topology, compactness follows directly from the fact that any open cover for a subset must include the entire subset itself (if non-empty) or be empty.

$\textbf{Connectedness:}$ A subset $A$ is connected in the trivial topology if the only way to express $A$ as the union of two disjoint open sets is to have one of them as $\emptyset$. For example, let's consider the set $[1, 3] \cup [4, 6]$ in the trivial topology on $\mathbb{R}$. In this topology, the only open sets are $\emptyset$ and $\mathbb{R}$. The set $[1, 3] \cup [4, 6]$ cannot be expressed as the union of two disjoint non-empty open sets. Therefore, it is connected in the trivial topology.

$\textbf{In summary:}$ $\textbf{Compactness:}$ All subsets of $X$ are compact. $\\$ $\textbf{Connectedness:}$ A subset is connected if it cannot be expressed as the union of two disjoint non-empty open sets in the trivial topology. For example, $[1, 3] \cup [4, 6]$ is connected in this topology.