7
$\begingroup$

I recently started studying Topology,and function defined from one topological space to another. I found in the Wikipedia ,where I was reading about subbase that 'continuity of a function need only be checked on a subbase of the range.'

That is, if $\mathcal B$ is a subbase for $Y$, a function  $f  : X \to Y$ is continuous if and only if  $f^{−1}(U)$ is open in $X$ for each $U \in\mathcal B$.

I tried to prove it by taking an open set in $Y$ and showing that its inverse is also open if $f$ is continuous, after that I studied the structure of $O$ and found that each $O$ is in fact union of finite intersections of elements of $\mathcal B$, and I got stuck, how to show that the inverse of unions of finite intersections of elements of $\mathcal B$ to be open?

May be it's very easy,but it's not striking my head, hope to get a help. Thank you.

$\endgroup$
6
$\begingroup$

You use the fact that, for a family of sets $(A_i\,:\,i\in I)$, $$f^{-1}\left(\bigcap_{i\in I}A_i\right)=\bigcap_{i\in I} f^{-1}(A_i)\\ f^{-1}\left(\bigcup_{i\in I}A_i\right)=\bigcup_{i\in I} f^{-1}(A_i)$$

Let's prove the first one. \begin{align}x\in f^{-1}\left(\bigcap_{i\in I}A_i\right)&\iff f(x)\in\bigcap_{i\in I}A_i\iff \forall i\in I,\ f(x)\in A_i\\&\iff \forall i\in I,\ x\in f^{-1}(A_i)\iff x\in \bigcap_{i\in I} f^{-1}(A_i)\end{align} The other one is similar. Once you know that, you're done: preimage of an open set is union of finite intersections of preimages of elements of the subbasis, thus union of finite intersections of open sets, thus open.

$\endgroup$
  • $\begingroup$ Thanks for the proof, can you give any hint about the converse part? $\endgroup$ – hiren_garai Mar 17 '17 at 6:13
  • $\begingroup$ That was for the part "preimage of prebase open$\implies$ continuous". The converse is trivial, because the elements of a prebase are always open. $\endgroup$ – user228113 Mar 17 '17 at 7:26
4
$\begingroup$

You can check it directly using the fact that the open sets are unions of finite intersections of the subbasis, but here is an approach which is also useful for other contexts (like measure theory, where measurable sets are not so easily described).

Firstly, we let the definition of topology generated by a subbasis be the natural one: it is the smallest topology containing the elements of the subbasis. It should be an easy exercise to check that this is equivalent to whatever definition you have.

Let $f: (X,\tau) \to (Y,\tau')$ be such that $f^{-1}(S)$ is open for every $S$ an element of the subbasis $\mathcal{S}$ of $(Y,\tau')$. Note now that it suffices to show that the set $$\mathcal{O}:=\{A \subset Y \mid f^{-1}(A) \in \tau\}$$ is a topology on $Y$. Since it contains the elements of $\mathcal{S}$ by hypothesis, we will have that $\tau' \subset \mathcal{O}$ by definition of subbasis. Consequently, $f$ is continuous.

But that $\mathcal{O}$ is a topology follows readily from the equations involving $f^{-1}$ and intersections/unions, and the fact that $\tau$ is a topology.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.