basis of open sets for the topological space of pointwise convergence

Question

Let $X= [0,1]^{[0,1]} = \{ f : f: [0,1] \rightarrow [0,1]\}$ be endowed with the topology $T$ of pointwise convergence.

Write a basis of open sets for the topological space $(X,T)$. Explain why this is a compact Hausdorff space.

Let $\epsilon >0$.

From reading some online notes, an open set in $(X,T)$ is of the form $O = \{f \in X : |f(x_i)-y_i|< \epsilon, i=1, \cdots , n \}$, where $x_i, \cdots, x_n \in [0,1] , y_i, \cdots, y_n \in [0,1]$. And a base is given by all finite intersections of sets of this form.

But I'm not sure why this is how open sets are defined. Why do we choose $x_i, \cdots, x_n$ and $y_i, \cdots, y_n$?

Also, why is a base the finite intersection of sets of the form?

Thanks in advance!

Answer 1

Pointwise convergence topology due to Wikipedia is the same as product topology, when we use nets instead of only sequences. Your question is in fact about product topology (or perhaps said equivalence).

Given arbitrary set of idicies $I$ and topological spaces $(X_i)_{i\in I}$ we define topology on the product $\prod_{i\in I} X_i$ as smallest topology in which all projections $p_j:\prod_{i\in I} X_i \to X_j$ defined as $p_j(x) = x_j$ are continuous. So every set of the form $p_j^{-1}(B_j)$ must be open in product, where $B_j$ is open set in $X_j$. It is sufficient to assume that $B_j$ is basis set in $X_j$.

If we take all $X_i$ to be the same, we could write $\prod_{i\in I} X = X^I$. Now elements of the product can be considered functions $f:I \to X$.

Take $X = I = [0, 1]$. As said before, $p_x^{-1}(B)$ must be open in product topology, where $B = \{y \in [0, 1] : |y - y_0| < \varepsilon \}$ for some appropriate constants $y_0, \varepsilon$. Notice that $$p_x^{-1}(B) = \{ f \in [0, 1]^{[0, 1]} : |f(x) - y_0| < \varepsilon \}. $$ The sets you are interested in are finite intersections of sets of in the above form and are called Cylinder sets. They need to be open because of our condition of projections being continuous and the fact that finite intersection of open sets is open. Because we were interested in smallest topology in which projections are continuous, we define basis to be exactly the cylinder sets. By definition they form $\pi$-system which covers the product space $[0, 1]^{[0, 1]}$, so we can actually define topology by naming such set family a basis. Set is open iff it is union of basis sets.

So chosen $x_1, x_2, \ldots, x_n, y_1, y_2, \ldots, y_n$ are really arbitrary (the standard basis consists of all possible choices of them, as well as all possible choices of finite $n$).

By Tychonoff's theorem, product of compact spaces is compact. Product of Hausdorff spaces is also Hausdorff, wchich can be checked directly using basis sets: if $f \neq g$, then $(\exists x) f(x) \neq g(x)$. Because all spaces were Hausdorff, we can take basis sets $U, V$ in $[0, 1]$ such that $f(x) \in U, g(x) \in V$ and $U \cap V = \emptyset$. We have $p_x^{-1}(U) \cap p_x^{-1}(V) = \emptyset$.

Note how in product topology we don't care if set of indices was topological space at all. The same sort of applies in pointwise convergence: functions must converge in every point separately, there is no notion of "close points" converging with "similar speed". This is how I think about it, I don't feel qualified to talk about net convergence induced topology being equivalent to product topology.

Answer 2

I guess that using sequences, $\varepsilon$, $[0,1]$, choices of $x_i$, $y_i$, etc, just make things more and more difficult to understand. As @Esgeriath has already stated, thinking about the product topology would probably be more adequate.

First of all, notice that you don't need to take finite intersections of sets in $\mathcal{O}$. This family is already a basis for the topology.

Saying, for example, that \begin{equation*} |f(x) - y| < \varepsilon \end{equation*} is just one way of saying that $f \in \pi_x^{-1}(y - \varepsilon, y + \varepsilon)$, where $\pi_x(f) = f(x)$ is the $x$-th coordinate of $f$. Another way, is saying that $f(x) \in (y-\varepsilon, y+\varepsilon)$.

There, it would be nice to notice that the open intervals form a basis for the usual topology of $\mathbb{R}$. Those are sets that you "want" to be open. If those are open, you can be sure that \begin{equation*} f_n \rightarrow f \Rightarrow f_n(x) \rightarrow y. \end{equation*}

Whenever you have a family of sets that you would like to be open, you can generate a topology. Since $\mathcal{O}$ covers $\mathcal{R}$, all you have to do is take finite intersections to get a topology.

Now, think of \begin{equation*} \mathcal{C} = \left\{V_{x,y,\varepsilon}\,|\; x \in [0,1], y \in [0,1], \varepsilon > 0 \right\}, \end{equation*} where \begin{equation*} V_{x,y,\varepsilon} = \{f|\,f(x) \in (y-\varepsilon, y+\varepsilon)\}. \end{equation*} (By the way, it would be more didactic if the domain and co-domain were not the same...)

Taking the family of finite intersections of members of $\mathcal{C}$, you would get almost your $\mathcal{O}$. The difference would be using $\varepsilon_1, \dotsc, \varepsilon_n$. This would be trivially a basis because it would simply be finite intersections. But any open interval of radius bigger then $\varepsilon > 0$ is a union of open intervals of radius $\varepsilon$. Therefore, you can take $\varepsilon = \min(\varepsilon_1, \dotsc, \varepsilon_n)$ and still have a basis.

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By the way, in general it is very hard to talk about topology of "sequence convergence": Topology for convergent sequences .