Compacts: propriété de Borel-Lebesgue.
The definition of compact subsets of a metric space is as follows:
Définition: Soit \(E\) be a metric space. A subset \(A\subset E\) is compact from any open covering of \(A\) we can find a finite subcovering , i.e.
if \[A\subset \bigcup_{i\in \mathcal{I}} O_i\] with the \(O_i\) open sets, then there exists \(\mathcal{J}\subset \mathcal{I}\) a finite set such that \[A\subset \bigcup_{i\in\mathcal{J}} O_i\]
The equivalent Bolzano-Weierstrass sequential characterization is more commonly used in exercises:
Property: Let \(E\) be a metric space. \(A\subset E\) is compact if and only if from any sequence of elements of \(A\) we can extract a subsequence which converges in \(A\).
The characterization of finite-dimensional compact parts is also frequently used:
Property: Let \(E\) be a finite-dimensional vector space. The compact subsets of \(E\) are the bounded and closed subsets.
I find the following exercise interesting because it involves the definition of compactness:
Exercice: Let \(f:\mathbb{R} \to \mathbb{R}\) be a function that has a limit on the left and a limit on the right at any point. Show that the points of discontinuity of \(f\) are at most countable.
Show that the number of discontinuities of \(f\) on any segment \(I\subset\mathbb{R}\) is at most countable.
We can quantify the discontinuity at \(x\) by \[d(x)=\lvert l_x^+-f(x) \rvert + \lvert l_x^--f(x) \rvert\] où \(l_x^+=\lim_{\underset{t>x}{t\to x}} f(t)\) et \(l_x^-=\lim_{\underset{t < x}{t\to x}} f(t)\).
What is the maximum value of the discontinuity \(d(t)\) for \(t\in ]x-\alpha_x,x+\alpha_x[\setminus \{x\}\)?
\[I\subset \bigcup_{x\in I}]x-\alpha_x,x+\alpha_x[\]
We use the following notations: \[l_x^+=\lim_{\underset{t>x}{t\to x}} f(t)\] \[l_x^-=\lim_{\underset{t < x}{t\to x}} f(t)\] \[d(x)=\lvert l_x^+-f(x) \rvert + \lvert l_x^--f(x) \rvert\]
Recall that: \[ \begin{alignat*}{1} f\text{ is coninuous at }x &\Leftrightarrow l_x^+=f(x) \land l_x^-=f(x) \\ \end{alignat*}\] and therefore \[ \begin{alignat*}{1} f\text{ is not continuous at }x &\Leftrightarrow l_x^+\neq f(x) \lor l_x^-\neq f(x) \\ &\Leftrightarrow \lvert l_x^+- f(x) \rvert >0 \lor \lvert l_x^- - f(x)\rvert >0 \\ &\Leftrightarrow d(x) =\lvert l_x^+- f(x) \rvert + \lvert l_x^- - f(x)\rvert >0 \\ \end{alignat*}\]
In what follows, we use the word segment to stand for closed bounded interval.
Set of discontinuity points of \(f\) in a compact.
Let \(I\) a segment of \(\mathbb{R}\). \(I\) is a compact subset because it is bounded and closed.
Let \(x\in I\). We use the assumption that \(f\) admits a real limit on the left and right:
Let \(\epsilon>0\). \[ \begin{alignat*}{1} \exists \alpha_x>0\quad \forall t\in\mathbb{R},\quad &t\in ]x-\alpha_x,x[\Rightarrow \lvert f(t)-l_x^-\rvert \leqslant \epsilon \\ &t\in ]x,x+\alpha_x[\Rightarrow \lvert f(t)-l_x^+\rvert \leqslant \epsilon \\ \end{alignat*}\]
By considering the left and right limits at each \(t\): \[ \begin{alignat*}{1} \forall t\in ]x-\alpha_x,x[,\quad &l_t^-.l_t^+ \in \overline{[l_x^--\epsilon,l_x^++\epsilon]} = [l_x^--\epsilon,l_x^++\epsilon] \\ \end{alignat*} \] and thus \[ \begin{alignat*}{1} \forall t\in ]x-\alpha_x,x[,\quad d(t)=\lvert l_t^+- f(t) \rvert + \lvert l_t^- - f(t)\rvert&= \lvert l_t^+ -l_x^- + l_x^- - f(t) \rvert + \lvert l_t^- -l_x^- + l_x^- - f(t)\rvert \\ &\leqslant \lvert l_t^+ -l_x^-\rvert + \lvert l_x^- - f(t) \rvert + \lvert l_t^- -l_x^-\rvert + \lvert l_x^- - f(t)\rvert \\ &\leqslant 4\epsilon \end{alignat*} \]
Similarly, \[ \begin{alignat*}{1} \forall t\in ]x,x+\alpha_x[,\quad d(t)&\leqslant 4\epsilon \end{alignat*} \]
Let \(n\in\mathbb{N}^*\). Choose \(\epsilon=\frac{1}{4n}\). We can cover \(I\) using the above open sets: \[ I \subset \bigcup_{x\in I} ]x-\alpha_x,x+\alpha_x[ \]
From the latter open covering, we extract a finite subcovering: \[ I \subset \bigcup_{i=1}^p ]x_i-\alpha_{x_i},x_i+\alpha_{x_i}[ \]
According to the above, discontinuities \(d(x)>\frac{1}{n}\) can only be found in the centers of these open intervals, so there is a finite number of them, at most \(p\).
We have: \begin{alignat*}{1} f\text{ is not continuous at }x \in I&\Leftrightarrow d(x)\in\mathbb{R}^{+*}=\bigcup_{n\in\mathbb{N}^*}]\frac{1}{n},+\infty[ \end{alignat*} which proves that the set of discontinuities of \(f\) in \(I\) is at most countable, because it is a countable union of finite sets.
Set of discontinuity points of \(f\) in \(\mathbb{R}\).
We can conclude that the set of discontinuities points of \(f\) in \(\mathbb{R}\) is at most countable since \(\mathbb{R}\) is a countable union of segments (for instance \(\mathbb{R}=\bigcup_{n\in\mathbb{Z}}[n,n+1]\)) and a countable union of at most countable sets is at most countable.