Is it necessary that Bases for every topology must exist?

Question

I was studying about continuous maps of topological spaces where I find an example

Let $f:X\to Y$, where $X$ and $Y$ are topological spaces and $\beta$ be bases for the topology on $Y$. Suppose for each member $B \in \beta$ then $f^{-1}[B]$ is an open subset of $X$. then $f$ is continuous function.

Here I'm going to add more information

For let $H$ be and open subset of $Y$ then $$f^{-1}[H] = f^{-1}[\underset{i}{\cup}B_i]=\underset{i}{\cup}f^{-1}[B_i]$$ where $B_i$ are members of $\beta$ and so $f^{-1}[B_i]$ is union of open sets is open. So $f^{-1}[H]$ is open in $X$. Hence $f$ is continuous map.

What I got if $Y$ has bases then function $f$ is must continuous.

My question is that, will every topology has bases, because if every topology has bases then function $f$ will must be continuous.

Answer 1

Every topological space $(X,\tau)$ has a base, namely $\tau$ is a base for $\tau$.

Just recall the definition:

A collection of open subsets $B\subseteq\tau$ is a base for $\tau$ if every element of $\tau$ is a union of elements of $B$

Do you see that $\tau$ trivially satisfies this definition?

Answer 2

The point of the theorem you quoted is laziness, not whether $Y$ has a base or not, which is independent of $f$ and $X$.

If you want to show $f: X \to Y$ is continuous, you must check by definition that $f^{-1}[O]$ is open in $X$ for any open subset $O$ of $Y$. But the theorem says that you can be lazy, and pick any base $\mathcal{B}$ for the topology of $Y$ that you like and only show that $f^{-1}[B]$ is open for $B \in \mathcal{B}$ instead (which you hope is less work, think of the open rational intervals in $\mathbb{R}$ as a base vs all open subsets), and be done with the continuity proof.

As noted, you could have chosen $\mathcal{T}_Y$ as a base and have no work-reduction, or $\mathcal{T}_Y\setminus\{\emptyset\}$ (also no work-reduction really as $f^{-1}[\emptyset] = \emptyset$ is trivially open anyways), but it's often applied when your favourite base is actually easier to work with.

If you really want to be lazy, use subbases for $Y$ instead (this is also a theorem, with almost identical proof as for bases).