I know the following definition of coarser topologies:

If $\tau_1$ and $\tau_2$ are two topologies on $X$, we say $\tau_1$ is coarser than $\tau_2$ if $\tau_1\subseteq\tau_2$.

In my book about topology there is an alternative way to describe coarser topologies:

$\tau_1$ coarser than $\tau_2 \iff \forall x\in X \, \forall B^1\in\mathcal{B}^1(x): \exists B^2\in \mathcal{B}^2(x):B^2\subseteq B^1$

where $\forall x\in X$, $\mathcal{B}^1(x),\mathcal{B}^2(x)$ are neighborhood bases for $\tau_1$ and $\tau_2$

I do not see the relation between coarser topologies and different neighborhood bases, may you could help me with that.


The below proof for the $\Leftarrow$ direction follows this ProofWiki proof.

The idea is that, given any basis element $B^1$, we can find a neighbourhood $B^2 \in \mathcal B^2(b)$ for each $b \in B^1$ since in this case $B^1 \in \mathcal B^1(b)$. $B^1$ will be the union of these neighborhoods. Then the neighborhood bases for $\tau_1$ are contained in $\tau_2$, hence so is $\tau_1$ itself.

So for each $b \in B^1$, $U_b = \bigcup \{B^2 \in \mathcal B^2(b): B^2 \subseteq B^1\}$ is non-empty, contained in $B^1$, and open in $\tau_2$.

It is clear that $B^1 \subseteq \bigcup\limits_{b \in B^1} U_b$; it is equally clear that the converse inclusion holds. Thus $B^1 \in \tau_2$.

We see that the trick is to apply the condition to every element of a (neighborhood basis) element $B^1$ of $\tau_1$. The approach taken above avoids using the Axiom of Choice by simply taking all valid $B^2$.

For the converse, $\Rightarrow$ direction, let $\mathcal B^1(x)$ be a neighbourhood basis around $x$ for $\tau_1$, and let $\mathcal B^2(x)$ be one for $\tau_2$.

Now every $B^1$ is an open $\tau_2$-neighbourhood around $x$ as well. By the definition of neighbourhood basis, there must be a $B^2 \in \mathcal B^2(x)$ such that $B^2 \subseteq B^1$.

Thus indeed the two are equivalent.

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  • $\begingroup$ Thank you, do you also have an idea for the other direction $=>$ ? $\endgroup$ – Voyage Apr 16 '13 at 9:01

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