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I'm having a lot of trouble with the following question:

Let $X$ be the set of of all functions that map $[0, 1]$ to itself. For each subset $A$ $\subset$ $[0, 1]$ let $B_A$ := {$f$ $\in$ $X$ | $f(x) = 0$ $\forall$x $\in$ A}. Prove that $\beta$ := {$B_A$ | $A$ $\subset$ $[0, 1]$} is a basis for a topology on X.

I think I'm having troubles identifying what the elements of each set actually are, which in turn is making it very difficult for me to see how this is a basis. For instance, if I take the constant function $f(x)$ = $1/2$ whose domain is $[0,1]$, certainly that function is in $X$, but I do not see how it is in the collection of basis elements.

I am very new to topology and really only have an understanding on the definitions of a topology and a basis. Thanks for any and all help!

1 Answers 1

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The basis element that contains the constant function $1/2$ is $B_\varnothing$, the set of all functions $f \in X$ that are nonzero on $[0,1]$.

To prove $\beta$ is a basis, we need to show

  1. it covers $X$ ,and
  2. for any $B_{A_1}$ and $B_{A_2}$ and any $f \in B_{A_1} \cap B_{A_2}$ there exists $B_{A_3}$ such that $x \in B_{A_3} \subset B_{A_1} \cap B_{A_2}$.

1. For any function $f$ let $A := \{x \in [0,1] : f(x)=0\}$. Then $f \in B_A$.

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2. Take $A_3 = A_1 \cup A_2$. Then $B_{A_3} = B_{A_1} \cap B_{A_2}$

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    Ok, so I've indeed misunderstood the set $B_A$. If we take A = {1/2}, $B_A$ is the set of all function such that 1/2 goes to 0, but not necessarily anything else. Does that sound correct? – 2017-01-23
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    @SenoritaChad That's correct. – 2017-01-23
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    Ok, could you also verify my understanding of the first part? To show $\beta$ covers $X$, we define the set $A$ for $f$ such that $B_A$ will contain functions of $X$ which are $0$ exactly when $f$ is $0$, so obviously $f$ is contained in $B_A$? I'm still trying to make sense of the second part... – 2017-01-23
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    @SenoritaChad That looks fine. – 2017-01-23