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.

  • 2
    $\begingroup$ I think the collection of all open sets would be a basis, so yes every topology has a basis. $\endgroup$ – Trevor Norton Apr 13 '18 at 14:02
  • $\begingroup$ Related: math.stackexchange.com/a/193470/291063 $\endgroup$ – Trevor Norton Apr 13 '18 at 14:03
  • $\begingroup$ Okay thanks, I got that. $\endgroup$ – Syed Muhammad Asad Apr 13 '18 at 14:17
  • $\begingroup$ If $X=\emptyset$ then the only topology on $X$ is $T=\{\emptyset\}$ and there are two bases for $T, $ which are $T$ and $\emptyset.$ $\endgroup$ – DanielWainfleet Apr 13 '18 at 20:45

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?

  • $\begingroup$ Thanks for response $\endgroup$ – Syed Muhammad Asad Apr 13 '18 at 14:17

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).

  • 1
    $\begingroup$ Exactly. A concrete example is when you use the $\epsilon$--$\delta$ definition of continuity in Euclidean space (or any metric space). You don't prove that the inverse image of every open set is open; you only prove that the inverse image of every $\epsilon$-ball is open. But this is sufficient because $\epsilon$-balls form a basis for the topology induced by the metric. $\endgroup$ – Robert Bell Apr 13 '18 at 14:47

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