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My professor wrote this:

"To study some topological property, we can always, without loss of generality, only focus on a basis for the topology."

Can someone explain this, and maybe give a simple example? I try an example below: If you want to prove some function $f\colon X \to Y$ is continuous, you don't have to take an arbitrary open set in $Y$ and show that its pre-image is open in $X$. Instead, you can take a basis element of $Y$'s topology and show its pre-image is open in $X$.


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Your example is a good one. You could also think about checking whether a point is in the closure of a subset of a topological space. – Dylan Moreland Mar 29 '12 at 3:02
How can one show the projection map $\pi:X\times Y \to X$ is an open map? – Abhishek Parab Mar 29 '12 at 3:04
If you can show the property holds for basic open sets, then the property will follow for arbitrary open sets, since every open set is a union of basic open sets, the inverse image of a union is the union of the inverse images, and the union of open sets is open. – Arturo Magidin Mar 29 '12 at 3:06
Often it is easier to define some mathematical object on a basis of the space and then extend the structure to the whole space. One example that comes to mind is the definition of an affine scheme, or more generally the definition of sheaves on a base. – Rankeya Mar 29 '12 at 3:19
haha! Abhishek Parab, this is exactly the proposition I was trying to show, which caused me to look into this stuff about bases. – Ryan M Mar 29 '12 at 3:20

Sure. The key idea here is that any open set can be written as a union of elements of your basis.

Let's take not just any open set in $Y$, but a basis element $B$. It's enough to show that $f^{-1}(B)$ is open for any such $B$. Then if $U=\bigcup B_i$ is any open set, $f^{-1}(U)=\bigcup f^{-1}(B_i)$ is a union of open sets, hence open.

Another example: A set is compact if any cover by basis elements has a finite subcover.

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Thank you very much! – Ryan M Mar 29 '12 at 3:13
And for the compactness example, we can even use subbases, as for continuity; not for openness of maps or closure though. – Henno Brandsma Mar 29 '12 at 20:38

The idea here is that taking unions is often a harmless procedure that can commute with different operations. If we can prove that some property holds on the basis for a topology, we can sometimes extend it to all open sets by taking unions.

In your example, suppose that $B$ is a basic open set, and $f^{-1}(B)$ is open in $X$. Since every open set $U$ can be written as a union $\bigcup_{i} B_{i}$ of open sets, then

$$f^{-1}(U) = f^{-1}(\bigcup_{i} B_{i}) = \bigcup_{i} f^{-1}(B_{i})$$

Since each $f^{-1}(B_{i})$ is open in $X$, their union $\bigcup_{i} f^{-1}(B_{i})$ is also open in $X$, so that $f$ is continuous.

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