8.10 Non-first-countable spaces[2BM]

E268

[0MM] 1 Prerequisites:2,265.Difficulty:*.Let \(Ω\) be a non-empty set; let’s consider \(X={\mathbb {R}}^Ω\).

  1. Let

    \[ U^ f_{E,𝜌}=\{ g∈ X, ∀ x∈ E, |f(x)-g(x)|{\lt}𝜌\} \]

    where \(f∈X\), \(𝜌{\gt}0\) and \(E⊂ Ω\) is finite. Show that the family of these \(U^ f_{E,𝜌}\) satisfies the requirements of 2, and is therefore a base for a topology \(𝜏\) (Hint: use 2). This topology is the product topology of topologies of \(ℝ\).

    In particular for each \(f\in X\) the sets \(U^ f_{E,𝜌}\) are a fundamental system of neighborhoods.

  2. Check that the topology is \(T_ 2\).

  3. Note that \(X\) is a vector space, and show that the “sum” operation is continuous, as an operation \(X× X→ X\); to this end, show that if \(f,g\in X,h=f+g\), for every neighborhood \(V_ h\) of \(h\) there are neighborhoods \(V_ f,V_ g\) of \(f,g\) such that \(V_ f+V_ g\subseteq V_ h\).

  4. Given \(B_ i⊂ℝ\) open and non-empty, one for each \(i∈Ω\), show that \(∏_ i B_ i\) is open if and only if \(B_ i=ℝ\) except at most finitely many \(i\).

Hidden solution: [UNACCESSIBLE UUID ’0MN’]

E268

[2BP] Prerequisites:265,1,261.Difficulty:*.Let \(Ω\) be an infinite uncountable set ; consider \(X=ℝ^Ω\) with the topology \(𝜏\) seen in 1.

  1. Show that every point in \((X,𝜏)\) does not admit a countable fundamental system of neighborhoods.

  2. Setting

    \begin{equation} C{\stackrel{.}{=}}\{ f∈ X, f(x)≠ 0 \text{~ for at most countably many ~ } x∈Ω\} \label{eq:C} \end{equation}
    269

    show that \(\overline C=X\);

  3. and that if \((f_ n)⊂ C\) and \(f_ n→ f\) pointwise then \(f∈ C\).

  4. Let \(I\) be the set of all finite subsets of \(Ω\), this is a filtering set if sorted by inclusion; consider the net

    \[ 𝜑:I→ X\quad ,𝜑(A) = {\mathbb 1}_ A \]

    then \(∀ A∈ I, 𝜑(A)∈ C\) but

    \[ \lim _{A∈ I} 𝜑(A) = {\mathbb 1}_ X∉ C\quad . \]

Hidden solution: [UNACCESSIBLE UUID ’2BQ’]

E268

[0MP]Difficulty:*.We restrict the topology described in the previous example to the set \(Y=[0,1]^{[0,1]}\) (that is, we restrict \(ℝ\) to \([0,1]\), and set \(Ω=[0,1]\)). Find a sequence \((f_ n)⊂ Y\) that does not allow a convergent subsequence.

Hidden solution: [UNACCESSIBLE UUID ’0MQ’]

Let’s recall the definition 252: a space \(X\) is ”compact by coverings” if, for every \((A_ i)_{i∈ I}\) family of open such that \(⋃_{i∈ I}A_ i= X\), there is a finite subfamily \(J⊂ I\) such that \(⋃_{i∈ J}A_ i= X\). The Tychonoff theorem shows that this space \(Y\) is ”compact by coverings”. This exercise shows you instead that \(Y\) it is not ”compact by sequences”.