How $f^{-1}(B_{\alpha})$ is open? 
i'm having some confusion in the above theory-


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*$f^{-1}(V)$ is open if $f^{-1}(B_{\alpha})$ is open.But how  $f^{-1}(B_{\alpha})$ is open?(this we have to show as said in the second line.)

 A: The point is that you suppose (or prove) that all $f^{-1}[B_\alpha]$ are open and then $f^{-1}[O]$ is open for all open $O$, so $f$ is continuous. If you already know $f$ is continuous, then of course all $f^{-1}[B_\alpha]$ are open, a fortiori (all open sets have open inverse images, so in particular all base elements have).
So to show continuity of $f$ you only have to show (the weaker statement, a priori) that the inverse images under $f$ of the elements of your favourite base or subbase for $Y$ are open in $X$; this can simplify some arguments.
A: The text explains how you can prove continuity of $f$ using the notions of basis and subbasis.
First, the text says "to prove continuity of $f$, it suffices to show that the inverse image of ever basis element is open" and the next sentence proves that claim. To be precise, that claim is: if every $f^{-1}(B_\alpha)$ is open, then $f$ is continuous. So there you assume that every $f^{-1}(B_\alpha)$ is open and conclude that $f^{-1}(V)$ is open.
Second, the text says "to prove continuity of $f$, it will even suffice to show that the inverse image of every subbasis element is open", again followed by a sentence that proves that claim.
