[proofplan] The proof is a direct unpacking of the definitions. Since $\tau$ is a topology on $X$, the defining axioms give both $\varnothing \in \tau$ and $X \in \tau$. Hence every element of the set $\{\varnothing, X\}$ belongs to $\tau$, so $\{\varnothing, X\} \subset \tau$. Finally, the [indiscrete topology](/page/Indiscrete%20Topology) on $X$ is precisely $\{\varnothing, X\}$, and the definition of coarser topology is inclusion of topologies. [/proofplan] [step:Extract the mandatory open sets from the topology axioms] Since $\tau$ is a topology on $X$, the definition of a topology on $X$ requires that both the empty subset and the whole underlying set belong to $\tau$. Therefore \begin{align*} \varnothing \in \tau \end{align*} and \begin{align*} X \in \tau. \end{align*} [guided] We use only the first containment axiom in the definition of a topology. The symbol $\tau$ denotes a collection of subsets of $X$, and saying that $\tau$ is a topology on $X$ means, among other axioms, that $\tau$ contains the empty subset of $X$ and the whole set $X$ itself. Thus the hypotheses already give \begin{align*} \varnothing \in \tau \end{align*} and \begin{align*} X \in \tau. \end{align*} No separation, countability, compactness, or other topological property is being used here; this is part of what it means for $\tau$ to be a topology. [/guided] [/step] [step:Convert the two memberships into inclusion of the indiscrete collection] Let $U$ be an arbitrary element of $\{\varnothing, X\}$. By the definition of a two-element listed set, either $U = \varnothing$ or $U = X$. In the first case, $U \in \tau$ because $\varnothing \in \tau$. In the second case, $U \in \tau$ because $X \in \tau$. Hence every element of $\{\varnothing, X\}$ belongs to $\tau$, so \begin{align*} \{\varnothing, X\} \subset \tau. \end{align*} This argument also covers the case $X=\varnothing$, where the listed set $\{\varnothing, X\}$ is just $\{\varnothing\}$ by extensionality of sets. [/step] [step:Identify the inclusion as coarseness of the indiscrete topology] By definition, the indiscrete topology on $X$ is the collection \begin{align*} \{\varnothing, X\}. \end{align*} By definition, a topology $\tau_0$ on $X$ is coarser than a topology $\tau_1$ on $X$ when $\tau_0 \subset \tau_1$. Applying this definition with $\tau_0=\{\varnothing, X\}$ and $\tau_1=\tau$, the inclusion proved above says exactly that the indiscrete topology on $X$ is coarser than $\tau$. Since $\tau$ was an arbitrary topology on $X$, the indiscrete topology is coarser than every topology on $X$. [/step]