Analysis Theorems

3 Topology of \(\RR \)

Lemma 3.1. An infinite subset \(X\) of a compact Hausdorff space \(Y\) has a limit point.

Proof. If not, \(X\) is closed, and is a countable compact discrete space, which is a contradiction. □

Proposition 3.2 (Baire’s Category Theorem). A locally compact Hausdorff space or complete metric space \(X\) is a Baire space.

Proof. First let’s treat locally compact Hausdorff. Let \(X_n\) be closed sparse subsets of \(X\), and \(U_0\) an open set in \(X\). We can inductively produce \(\bar {U_i} \subset U_{i-1}\) that avoid \(X_1,\dots ,X_i\) by regularity. But then \(\cap _i\bar {U_i}\) is nonempty, so there is a point that avoids all \(U_i\). Similarly in the complete metric space case, we can choose the \(U_i\) in the same way with the condition that the diameter of \(U_i\) is less than \(1/n\). Then again the intersection must be a single point by completeness. □

Theorem 3.3. The unit cube \(I^n\) is compact.

Proof. Suppose we have a covering with no finite subcover. Then we can divide the unit cube into \(2^n\) pieces of half the size, and at least one of these must have the same property. We can keep doing this, getting a chain of cubes contained within one another with diameter approaching \(0\) such that this cover has no finite subcover for any of these. But the intersection of these has diameter \(0\), but contains exactly \(1\) point by taking the limiting point of a sequence of points in each of the cubes in the chain. Taking an open set surrounding this point, we get a contradiction as infinitely many cubes must be in the open set. □

Corollary 3.4 (Heine-Borel). A subset of \(\RR ^n\) is compact iff it is closed and bounded.

Proof. If a subset is compact, it is closed as \(\RR ^n\) is Hausdorff, and bounded by looking at the cover given by open balls around the origin. Conversely if a subset is closed and bounded, it is a closed subset of a large unit cube, hence is compact by Theorem 3.3. □

Corollary 3.5 (Bolzano-Weierstrass). A subset of \(\RR ^n\) is compact iff any sequence has a convergent subsequence.

Proof. ‘ Suppose any sequence has a convergent subsequence. Then the subset must be bounded or else we could take an increasingly large sequence. It must also be closed or else we could take points in decreasingly small neighborhoods of a limit point. The converse follows from Lemma 3.1. □

Lemma 3.6 (Minimum-Maximum Theorem). A map \(f:S \to \RR \) from compact \(S\) has a min and max.

Proof. The image is compact, hence closed and bounded, so attains its supremum and infimum. □

Theorem 3.7. All norms on finite dimensional \(\RR \)-vector spaces are equivalent.

Proof. We will show every norm is equivalent to \(\Vert \cdot \Vert _\infty \). To do this, let \(\Vert \cdot \Vert \) be any norm, and we will show it is continuous with respect to \(\Vert \cdot \Vert _\infty \). In particular, if \(M = \max _i\Vert \delta _i \Vert \) we have \(\Vert x\Vert \leq \sum _1^n\Vert x_i\delta _i\Vert \leq M\sum _1^n|x_i| \leq Mn\Vert x\Vert _\infty \), so we have \(|\Vert x \Vert - \Vert a \Vert | \leq \Vert x-a\Vert \leq Mn\Vert x-a \Vert _\infty \) so indeed it is continuous.

The set \(X = \{x|\Vert x \Vert _\infty = 1\}\) is compact, so its image from \(\Vert \cdot \Vert \) by Lemma 3.6 has a minimum \(m\) and a maximum \(M\). Thus \(m\) and \(M\) give the bounds we need for equivalence of norms. □

Lemma 3.8. There is no retraction from \(D^n \to S^n\).

Proof. Look at \(H^{n-1}\). □

Proposition 3.9 (Intermediate Value Theorem). A map from \(D^n\) to itself that is the identity on the boundary is surjective.

Proof. If not, we can deformation retract away from the missing point to the boundary, to yield a retraction of \(D^n\) onto \(S^{n-1}\), contradicting Lemma 3.8. In the case \(n=1\) this is looking at the connected component. □

Proposition 3.10 (Brouwer Fixed Point Theorem). A map \(f:D^n \to D^n\) has a fixed point.

Proof. If not, we can consider the line going from \(f(x)\) to \(x\) and produce a retraction \(g\) that is the intersection (on the \(x\) side) of this line with the boundary, contradicting Lemma 3.8. □

Theorem 3.11. For any embedding \(i:D^k \to S^n\), \(S^n-i(D^k)\) has trivial \(\tilde {H_*}\).

Proof. We induct on \(k\), for \(k =0\) we just have \(S^n-i(D_0) = \RR ^n\). For the induction step, we replace \(D^k\) with \(I^k\). Now we split \(I^k\) into two halves \(I^{k-1}\times [0,\frac {1}{2}],I^{k-1}\times [\frac {1}{2},1]\) and by Mayer-Vietoris and induction we have \(\tilde {H}_*(S^n-I^k) \hookrightarrow \tilde {H}_*(I^{k-1}\times [0,\frac {1}{2}])\oplus \tilde {H}_*(I^{k-1}\times [\frac {1}{2},1])\) is an isomorphism. If there were a nontrivial (reduced) cycle \(\alpha \) in \(H_*(S^n-I^k)\) it would land nontrivially in one of the two summands on the right, and we could repeatedly cut these up into smaller intervals, in the limit landing in \(H_*(S^n-I^{k-1})\) in which we know it would be a boundary of a chain \(\beta \). But then as \(\beta \) is compact and covered by our cuts, it is in one of the cuts, a contradiction. □

The next theorem has the Jordan Curve Theorem as a special case:

Theorem 3.12. For any embedding \(i:S^k \to S^n,n>k\) we have \(\tilde {H}_i (S^n-i(S^k)) = \ZZ ^{\delta _{i,n-k-1}}\).

Proof. We induct on \(k\), for \(k=0\) we have \(S^n-i(S^0) \simeq S^{n-1}\). Now to induct we split \(S^k\) into to disks and use Mayer-Vietoris and Theorem 3.11 to give \(H_{m-1}(S^n-i(S^k))\cong H_{m}(S^n-i(S^{k-1})).\) □

Theorem 3.13 (Invariance of Domain). An injective map \(\RR ^n \to \RR ^n\) is open.

Proof. By Theorem 3.12, the boundary of an embedding \(D^n \to \RR ^n\) separates \(\RR ^n\) into two components, and since the boundary is closed, the interior of \(D^n\) is an connected component of this separation that is open in \(\RR ^n\). □

Lemma 3.14 (Lebesgue Number Lemma). Any open cover of a compact metric space \(X\) has a \(\delta \) (Lebesgue number) so that every \(\delta \) ball is in some open set.

Proof. Take a finite subcover \(U_1, \dots , U_n\), and consider the map \(f:X \to \RR \), \(f(x) = \sum d(x,X-U_i)\). Now this function attains a minimum \(\delta \), but \(\delta \) cannot be \(0\) as the \(U_i\) cover \(X\). Then \(\delta \) is a Lebesgue number. □

Proposition 3.15. A continuous map \(f:S \to T\) where \(S,T\) are metric spaces, and \(S\) is compact is uniformly continuous.

Proof. Take the preimages of \(\ee \) balls around each point in \(T\) to make a cover of \(S\). The Lebesgue number of this cover is a uniform \(\delta \) for this \(\ee \). □

Theorem 3.16 (Uniform Limit Theorem). A uniform limit of continuous functions to a metric space is continuous.

Proof. Let the sequence be \(f_n\) and the limit be \(f\). We need that for any \(\ee >0\) and any \(x \in X\) there is a neighborhood \(x \in U_\ee \) such that \(d(f(x),f(y))<\ee \) when \(x,y \in U_\ee \). We can choose large enough \(n\) so we have \(\forall x,d(f(x),f_n(x))<\ee \), and a neighborhood \(U\) satisfying \(\ee \) for \(f_n\) Then we have in \(U\):

\[ d(f(x),f(y)) \leq d(f(x),f_n(x))+d(f_n(x),f_n(y))+d(f_n(y),f(y)) < \ee +\ee +\ee =3\ee \]

so we are done. □

Theorem 3.17. If a sequence of continuous functions \(f_n\) from a compact space \(X\) to a metric space \(Y\) converge to a function \(f\), they uniformly converge to that function, which is continuous by Theorem 3.16.

Proof. For any \(\ee >0\), there is a neighborhood around each point \(x\in X\), \(U_x\) such that \(f(y)\) is within \(\ee /2\) of \(f(x)\) for any \(y \in U_x\). Then make a finite cover of these \(U_x\) and take the maximum \(N\) for each of the corresponding points such that \(f_N\) is within \(\frac {\ee }{2}\) for these points. Then by the triangle inequality every point simultaneously satisfies \(|f(x)-f_N(x)|\leq \ee \) for large enough \(N\). □

Theorem 3.18 (Ascoli). If \(f_n:X \to Y\) form a uniformly equicontinuous sequence of maps between compact metric spaces \(X\) and \(Y\), there is a uniformly convergent subsequence.

Proof. \(X\) has a countable dense subset \(x_n\), so starting with \(x_1\) we can look at \(f_i(x_1)\), which has a limit point by Lemma 3.1. This gives a subsequence which converges at \(x_1\). We can then inductively produce more subsequences for each \(x_n\) and take a diagonal sequence \(g_n\). Every compact metric space is complete by Lemma 3.1, so it suffices to show that \(g_n\) is uniformly Cauchy. To do so, fix \(\ee >0\), and cover \(X\) with finitely many neighborhoods around the \(x_i\) so each \(f_n\) varies at most \(\frac {\ee }{3}\) in the neighborhood. Each neighborhood contains an \(x_i\) so choose the maximum of the \(N\)s so for \(j\geq N\) the \(g_j(x_i)\) differ by at most \(\frac {\ee }{3}\). Then for any \(x \in X\), \(j,k>N\) we have \(d(g_j(x),g_k(x)) \leq d(g_j(x),g_j(x_i)) + d(g_j(x_i),g_k(x_i)) + d(g_k(x_i),g_k(x)) < \ee \). □