Criterion for Coarser/Finer Topologies

Theorem 1: Let $(X, \tau_1)$ and $(X, \tau_2)$ be topological spaces. Then $\tau_1$ is coarser than $\tau_2$ if and only if for each $x \in X$ and every $U \in \tau_1$ with $x in U$ there exists a $V \in \tau_2$ such that $x \in V \subseteq U$.

Theorem 1 merely states that if $\tau_1$ is a coarser (weaker) topology than $\tau_2$ on $X$, then the open neighbourhoods of $\tau_2$ are "smaller" than the open neighbourhoods of $\tau_1$.

• Proof: $\Rightarrow$ Suppose that $\tau_1$ is coarser than $\tau_2$. Then $\tau_1 \subseteq \tau_2$. Let $x \in X$ and let $U \in \tau_1$. Then $U \in \tau_2$. If we set $V = U$ then $x \in V \subseteq U$.

• $\Leftarrow$ Suppose that for every $x \in X$ and every $U \in \tau_1$ with $x \in U$ there exists a $V \in \tau_2$ such that $x \in V \subseteq U$.

• Let $A \in \tau_1$. For each point $x \in A$ we are guaranteed the exist of a set $B_x \in \tau_2$ such that $x \in B-x \subseteq A$. Indeed:

(1)

\begin{align} \quad A = \bigcup_{x \in A} B_x \end{align}

• Since arbitrary unions of $\tau_2$ open sets are $\tau_2$ open, the above equality tells us that $A$ is $\tau_2$ open, i.e., $A \in \tau_2$. Thus $\tau_1 \subseteq \tau_2$, i.e., $\tau_1$ is coarser/weaker than $\tau_2$. $\blacksquare$