index int64 1 367 | name stringlengths 17 29 | textbook stringclasses 10
values | header stringclasses 60
values | helper stringclasses 14
values | informal_stmt stringlengths 42 474 | formal_stmt stringlengths 66 545 | informal_proof stringlengths 40 3.85k | formal_proof stringclasses 1
value |
|---|---|---|---|---|---|---|---|---|
301 | Munkres_exercise_22_5 | Munkres | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology | Let $p \colon X \rightarrow Y$ be an open map. Show that if $A$ is open in $X$, then the map $q \colon A \rightarrow p(A)$ obtained by restricting $p$ is an open map. | theorem Munkres_exercise_22_5 {X Y : Type*} [TopologicalSpace X]
[TopologicalSpace Y] (p : X → Y) (hp : IsOpenMap p)
(A : Set X) (hA : IsOpen A) :
IsOpenMap (Set.MapsTo.restrict p A (p '' A) (Set.mapsTo_image p A)) := by
sorry | Let $U$ be open in $A$. Since $A$ is open in $X, U$ is open in $X$ as well, so $p(U)$ is open in $Y$. Since $q(U)=p(U)=p(U) \cap p(A)$, the set $q(U)$ is open in $p(A)$. Thus $q$ is an open map. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
302 | Munkres_exercise_23_3 | Munkres | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology | Let $\left\{A_{\alpha}\right\}$ be a collection of connected subspaces of $X$; let $A$ be a connected subset of $X$. Show that if $A \cap A_{\alpha} \neq \varnothing$ for all $\alpha$, then $A \cup\left(\bigcup A_{\alpha}\right)$ is connected. | theorem Munkres_exercise_23_3 {X α : Type*} [TopologicalSpace X]
{A : α → Set X}
(hAa : ∀ a, IsConnected (A a))
(A₀ : Set X)
(hA : IsConnected A₀)
(h : ∀ a, A₀ ∩ A a ≠ ∅) :
IsConnected (A₀ ∪ (⋃ a, A a)) := by
sorry | For each $\alpha$ we have $A \cap A_\alpha \neq \emptyset$, so each $A \cup A_\alpha$ is connected by Theorem 23.3. In turn $\left\{A \cup A_\alpha\right\}_\alpha$ is a collection of connected spaces that have a point in common (namely any point in $A)$, so $\bigcup_\alpha\left(A \cup A_\alpha\right)=A \cup\left(\bigcu... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
303 | Munkres_exercise_23_6 | Munkres | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology | Let $A \subset X$. Show that if $C$ is a connected subspace of $X$ that intersects both $A$ and $X-A$, then $C$ intersects $\operatorname{Bd} A$. | theorem Munkres_exercise_23_6 {X : Type*}
[TopologicalSpace X] {A C : Set X} (hc : IsConnected C)
(hCA : C ∩ A ≠ ∅) (hCXA : C ∩ Aᶜ ≠ ∅) :
C ∩ (frontier A) ≠ ∅ := by
sorry | Suppose that $C \cap B d A=C \cap \bar{A} \cap \overline{X-A}=\emptyset$. Then $C \cap A$ and $C \cap(X \backslash A)$ are a pair of disjoint non-empty sets whose union is all of $C$, neither of which contains a limit point of the other. Indeed, if $C \cap(X-A)$ contains a limit point $x$ of $C \cap A$, then $x \in C \... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
304 | Munkres_exercise_23_11 | Munkres | import Mathlib
open Filter Set TopologicalSpace Topology | Let $p: X \rightarrow Y$ be a quotient map. Show that if each set $p^{-1}(\{y\})$ is connected, and if $Y$ is connected, then $X$ is connected. | theorem Munkres_exercise_23_11 {X Y : Type*} [TopologicalSpace X] [TopologicalSpace Y]
(p : X → Y) (hq : IsQuotientMap p)
(hY : ConnectedSpace Y) (hX : ∀ y : Y, IsConnected (p ⁻¹' {y})) :
ConnectedSpace X := by
sorry | Suppose that $U$ and $V$ constitute a separation of $X$. If $y \in p(U)$, then $y=p(x)$ for some $x \in U$, so that $x \in p^{-1}(\{y\})$. Since $p^{-1}(\{y\})$ is connected and $x \in U \cap p^{-1}(\{y\})$, we have $p^{-1}(\{y\}) \subset U$. Thus $p^{-1}(\{y\}) \subset U$ for all $y \in p(U)$, so that $p^{-1}(p(U)) \s... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
305 | Munkres_exercise_24_3a | Munkres | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology | abbrev I : Set ℝ := Icc 0 1 | Let $f \colon X \rightarrow X$ be continuous. Show that if $X = [0, 1]$, there is a point $x$ such that $f(x) = x$. (The point $x$ is called a fixed point of $f$.) | theorem Munkres_exercise_24_3a (f : I → I) (hf : Continuous f) :
∃ (x : I), f x = x := by
sorry | If $f(0)=0$ or $f(1)=1$ we are done, so suppose $f(0)>0$ and $f(1)<1$. Let $g:[0,1] \rightarrow[0,1]$ be given by $g(x)=f(x)-x$. Then $g$ is continuous, $g(0)>0$ and $g(1)<0$. Since $[0,1]$ is connected and $g(1)<0<g(0)$, by the intermediate value theorem there exists $x \in(0,1)$ such that $g(x)=0$, that is, such that... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip |
306 | Munkres_exercise_25_9 | Munkres | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology | Let $G$ be a topological group; let $C$ be the component of $G$ containing the identity element $e$. Show that $C$ is a normal subgroup of $G$. | theorem Munkres_exercise_25_9 {G : Type*} [TopologicalSpace G] [Group G]
[IsTopologicalGroup G] (C : Set G) (hC : C = connectedComponent (1 : G)) :
∃ H : Subgroup G, (H : Set G) = C ∧ H.Normal := by
sorry | Given $x \in G$, the maps $y \mapsto x y$ and $y \mapsto y x$ are homeomorphisms of $G$ onto itself. Since $C$ is a component, $x C$ and $C x$ are both components that contain $x$, so they are equal. Hence $x C=C x$ for all $x \in G$, so $C$ is a normal subgroup of $G$. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
307 | Munkres_exercise_26_12 | Munkres | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology | Let $p: X \rightarrow Y$ be a closed continuous surjective map such that $p^{-1}(\{y\})$ is compact, for each $y \in Y$. (Such a map is called a perfect map.) Show that if $Y$ is compact, then $X$ is compact. | theorem Munkres_exercise_26_12 {X Y : Type*} [TopologicalSpace X] [TopologicalSpace Y]
(p : X → Y) (h : Function.Surjective p) (hc : Continuous p)
(hp : IsClosedMap p) (hp : ∀ y, IsCompact (p ⁻¹' {y}))
(hY : CompactSpace Y) : CompactSpace X := by
sorry | We first show that if $U$ is an open set containing $p^{-1}(\{y\})$, then there is a neighbourhood $W$ of $y$ such that $p^{-1}(W)$ is contained in $U$. Since $X-U$ is closed in $X$, $p(X-U)$ is closed in $Y$ and does not contain $y$, so $W=Y \backslash p(X \backslash U)$ is a neighbourhood of $y$. Moreover, since $X \... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
308 | Munkres_exercise_28_4 | Munkres | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology | def countably_compact (X : Type*) [TopologicalSpace X] :=
∀ U : ℕ → Set X,
(∀ i, IsOpen (U i)) ∧ ((univ : Set X) ⊆ ⋃ i, U i) →
(∃ t : Finset ℕ, (univ : Set X) ⊆ ⋃ i ∈ t, U i)
def limit_point_compact_strong (X : Type*) [TopologicalSpace X] :=
∀ U : Set X, Infinite U → ∃ x, ClusterPt x (𝓟 (U \ {x})) | A space $X$ is said to be countably compact if every countable open covering of $X$ contains a finite subcollection that covers $X$. Show that for a $T_1$ space $X$, countable compactness is equivalent to limit point compactness. | theorem Munkres_exercise_28_4 {X : Type*} [TopologicalSpace X]
(hT1 : T1Space X) : countably_compact X ↔ limit_point_compact_strong X := by
sorry | First let $X$ be a countable compact space. Note that if $Y$ is a closed subset of $X$, then $Y$ is countable compact as well, for if $\left\{U_n\right\}_{n \in \mathbb{Z}_{+}}$is a countable open covering of $Y$, then $\left\{U_n\right\}_{n \in \mathbb{Z}_{+}} \cup(X \backslash Y)$ is a countable open covering of $X$;... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip |
309 | Munkres_exercise_28_6 | Munkres | import Mathlib
open Filter Set TopologicalSpace Function
open scoped Topology | Let $(X, d)$ be a metric space. If $f: X \rightarrow X$ satisfies the condition $d(f(x), f(y))=d(x, y)$ for all $x, y \in X$, then $f$ is called an isometry of $X$. Show that if $f$ is an isometry and $X$ is compact, then $f$ is bijective and hence a homeomorphism. | theorem Munkres_exercise_28_6 {X : Type*} [MetricSpace X]
[CompactSpace X] {f : X → X} (hf : Isometry f) :
Continuous f ∧ Bijective f ∧ IsOpenMap f := by
sorry | Note that $f$ is an imbedding. It remains to prove that $f$ is surjective. Suppose it is not, and let $a \in f(X)$. Since $X$ is compact, $f(X)$ is compact and hence closed (every metric space is Hausdorff). Thus, there exists $\varepsilon>0$ such that the $\varepsilon^{-}$ neighbourhood of $a$ is contained in $X \back... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
310 | Munkres_exercise_29_4 | Munkres | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology BoundedContinuousFunction | abbrev I : Set ℝ := Icc 0 1 | Show that $[0, 1]^\omega$ is not locally compact in the uniform topology. | theorem Munkres_exercise_29_4 :
¬ LocallyCompactSpace (ℕ →ᵇ I) := by
sorry | Consider $\mathbf{0} \in[0,1]^\omega$ and suppose that $[0,1]^\omega$ is locally compact at $\mathbf{0}$. Then there exists a compact $C$ containing an open ball $B=B_\rho(\mathbf{0}, \varepsilon) \subset[0,1]^\omega$. Note that $\bar{B}=[0, \varepsilon]^\omega$. Then $[0, \varepsilon]^\omega$ is closed and contained i... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip |
311 | Munkres_exercise_30_10 | Munkres | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology | Show that if $X$ is a countable product of spaces having countable dense subsets, then $X$ has a countable dense subset. | theorem Munkres_exercise_30_10
{X : ℕ → Type*} [∀ i, TopologicalSpace (X i)]
(h : ∀ i, ∃ (s : Set (X i)), Countable s ∧ Dense s) :
∃ (s : Set (Π i, X i)), Countable s ∧ Dense s := by
sorry | Let $\left(X_n\right)$ be spaces having countable dense subsets $\left(A_n\right)$. For each $n$, fix an arbitrary $x_n \in X_n$. Consider the subset $A$ of $X$ defined by
$$
A=\bigcup\left\{\prod U_n: U_n=A_n \text { for finitely many } n \text { and is }\left\{x_n\right\} \text { otherwise }\right\} .
$$
This set is ... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
312 | Munkres_exercise_31_1 | Munkres | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology | Show that if $X$ is regular, every pair of points of $X$ have neighborhoods whose closures are disjoint. | theorem Munkres_exercise_31_1 {X : Type*} [TopologicalSpace X]
[RegularSpace X] [T1Space X]
(x y : X) (hxy : x ≠ y) :
∃ U V : Set X, IsOpen U ∧ IsOpen V ∧ x ∈ U ∧ y ∈ V ∧
closure U ∩ closure V = (∅ : Set X) := by
sorry | Let $x, y \in X$ be two points such that $x \neq y$. Since $X$ is regular (and thus Hausdorff), there exist disjoint open sets $U, V \subseteq X$ such that $x \in U$ and $y \in V$.
Note that $y \notin \bar{U}$. Otherwise $V$ must intersect $U$ in a point different from $y$ since $V$ is an open neighborhood of $y$, whic... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
313 | Munkres_exercise_31_3 | Munkres | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology | Show that every order topology is regular. | theorem Munkres_exercise_31_3 {α : Type*} [LinearOrder α] [TopologicalSpace α]
[OrderTopology α] : RegularSpace α := by
sorry | Let $X$ be an ordered set.
First we show that $X$ is a $T_1$-space. For $x \in X$ we have that
$$
X \backslash\{x\}=\langle-\infty, x\rangle \cup\langle x,+\infty\rangle
$$
which is an open set as an union of two open intervals. Therefore, the set $\{x\}$ is closed.
Step 2
2 of 3
Now to prove that $X$ is regular we use... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
314 | Munkres_exercise_32_2a | Munkres | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology | Show that if $\prod X_\alpha$ is Hausdorff, then so is $X_\alpha$. Assume that each $X_\alpha$ is nonempty. | theorem Munkres_exercise_32_2a
{ι : Type*} {X : ι → Type*} [∀ i, TopologicalSpace (X i)]
(h : ∀ i, Nonempty (X i)) (h2 : T2Space (Π i, X i)) :
∀ i, T2Space (X i) := by
sorry | Suppose that $X=\prod_\beta X_\beta$ is Hausdorff and let $\alpha$ be any index.
Let $x, y \in X_\alpha$ be any points such that $x \neq y$. Since all $X_\beta$ are nonempty, there exist points $\mathbf{x}, \mathbf{y} \in X$ such that $x_\beta=y_\beta$ for every $\beta \neq \alpha$ and $x_\alpha=x, y_\alpha=y$.
Since $... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
315 | Munkres_exercise_32_2c | Munkres | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology | Show that if $\prod X_\alpha$ is normal, then so is $X_\alpha$. Assume that each $X_\alpha$ is nonempty. | theorem Munkres_exercise_32_2c
{ι : Type*} {X : ι → Type*} [∀ i, TopologicalSpace (X i)]
(h : ∀ i, Nonempty (X i)) (h2 : NormalSpace (Π i, X i)) :
∀ i, NormalSpace (X i) := by
sorry | Suppose that $X=\prod_\beta X_\beta$ is normal and let $\alpha$ be any index.
Since $X$ is normal, it follows that $X$ is Hausdorff (or regular), which then implies that $X_\alpha$ is Hausdorff (or regular). This imples that $X_\alpha$ satisfies the $T_1$ axiom.
Now the proof that $X_\alpha$ satisfies the $T_4$ axiom i... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
316 | Munkres_exercise_33_7 | Munkres | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology | Show that every locally compact Hausdorff space is completely regular. | theorem Munkres_exercise_33_7 {X : Type*} [TopologicalSpace X]
(hX : LocallyCompactSpace X) (hX' : T2Space X) :
CompletelyRegularSpace X := by
sorry | $X$ is a subspace of a compact Hausdorff space $Y$, its one-point compactification. $Y$ is normal, and so by the Urysohn lemma $Y$ is completely regular. Therefore by corollary $X$ is completely regular. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
317 | Munkres_exercise_34_9 | Munkres | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology | Let $X$ be a compact Hausdorff space that is the union of the closed subspaces $X_1$ and $X_2$. If $X_1$ and $X_2$ are metrizable, show that $X$ is metrizable. | theorem Munkres_exercise_34_9
(X : Type*) [TopologicalSpace X] [CompactSpace X] [T2Space X]
(X1 X2 : Set X) (hX1 : IsClosed X1) (hX2 : IsClosed X2)
(hX : X1 ∪ X2 = univ) (hX1m : MetrizableSpace X1)
(hX2m : MetrizableSpace X2) : MetrizableSpace X := by
sorry | Both $X_1$ and $X_2$ are compact, Hausdorff and metrizable, so by exercise 3 they are second countable, i.e. there are countable bases $\left\{U_{i, n} \subset X_i \mid n \in \mathbb{N}\right\}$ for $i \in\{1,2\}$. By the same exercise it is enough to show that $X$ is second countable. If $X_1 \cap X_2=\emptyset$ both ... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
318 | Munkres_exercise_43_2 | Munkres | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology | Let $(X, d_X)$ and $(Y, d_Y)$ be metric spaces; let $Y$ be complete. Let $A \subset X$. Show that if $f \colon A \rightarrow Y$ is uniformly continuous, then $f$ can be uniquely extended to a continuous function $g \colon \bar{A} \rightarrow Y$, and $g$ is uniformly continuous. | theorem Munkres_exercise_43_2 {X : Type*} [MetricSpace X]
{Y : Type*} [MetricSpace Y] [CompleteSpace Y] (A : Set X)
(f : X → Y) (hf : UniformContinuousOn f A) :
∃ (g : closure A → Y),
UniformContinuous g ∧
(∀ x : A, g ⟨x, subset_closure x.property⟩ = f x) ∧
(∀ g' : closure A → Y, Conti... | Let $\left(X, d_X\right)$ and $\left(Y, d_Y\right)$ be metric spaces; let $Y$ be complete. Let $A \subset X$. It is also given that $f: A \longrightarrow X$ is a uniformly continuous function. Then we have to show that $f$ can be uniquely extended to a continuous function $g: \bar{A} \longrightarrow Y$, and $g$ is unif... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
319 | Pugh_exercise_2_26 | Pugh | import Mathlib
open Filter Real Function
open scoped Topology | Prove that a set $U \subset M$ is open if and only if none of its points are limits of its complement. | theorem Pugh_exercise_2_26 {M : Type*} [TopologicalSpace M]
(U : Set M) : IsOpen U ↔ ∀ x ∈ U, ¬ ClusterPt x (𝓟 Uᶜ) := by
sorry | Assume that none of the points of $U$ are limits of its complement, and let us prove that $U$ is open. Assume by contradiction that $U$ is not open, so there exists $p \in M$ so that $\forall r>0$ there exists $q \in M$ with $d(p, q)<r$ but $q \notin U$. Applying this to $r=1 / n$ we obtain $q_n \in U^c$ such that $d\l... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
320 | Pugh_exercise_2_32a | Pugh | import Mathlib
open Filter Real Function
open scoped Topology | Show that every subset of $\mathbb{N}$ is clopen. | theorem Pugh_exercise_2_32a (A : Set ℕ) : IsClopen A := by
sorry | 32. The one-point set $\{n\} \subset \mathbb{N}$ is open, since it contains all $m \in \mathbb{N}$ that satisfy $d(m, n)<\frac{1}{2}$. Every subset of $\mathbb{N}$ is a union of one-point sets, hence is open. Then every set it closed, since its complement is necessarily open. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
321 | Pugh_exercise_2_46 | Pugh | import Mathlib
open Filter Real Function
open scoped Topology | Assume that $A, B$ are compact, disjoint, nonempty subsets of $M$. Prove that there are $a_0 \in A$ and $b_0 \in B$ such that for all $a \in A$ and $b \in B$ we have $d(a_0, b_0) \leq d(a, b)$. | theorem Pugh_exercise_2_46 {M : Type*} [MetricSpace M]
{A B : Set M} (hA : IsCompact A) (hB : IsCompact B)
(hAB : Disjoint A B) (hA₀ : A ≠ ∅) (hB₀ : B ≠ ∅) :
∃ a₀ b₀, a₀ ∈ A ∧ b₀ ∈ B ∧ ∀ (a : M) (b : M),
a ∈ A → b ∈ B → dist a₀ b₀ ≤ dist a b := by
sorry | Let $A$ and $B$ be compact, disjoint and non-empty subsets of $M$. We want to show that there exist $a_0 \in A, b_0 \in B$ such that for all $a \in A, b \in B$,
$$
d\left(a_0, b_0\right) \leq d(a, b) .
$$
We saw in class that the distance function $d: M \times M \rightarrow \mathbb{R}$ is continuous. We also saw in cla... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
322 | Pugh_exercise_2_92 | Pugh | import Mathlib
open Filter Real Function Set
open scoped Topology | Give a direct proof that the nested decreasing intersection of nonempty covering compact sets is nonempty. | theorem Pugh_exercise_2_92 {α : Type*} [TopologicalSpace α] [T2Space α]
{s : ℕ → Set α}
(hcomp : ∀ i, IsCompact (s i))
(hne : ∀ i, (s i).Nonempty)
(hmono : ∀ i, s i ⊇ s (i + 1)) :
(⋂ i, s i).Nonempty := by
sorry | Let
$$
A_1 \supset A_2 \supset \cdots \supset A_n \supset \cdots
$$
be a nested decreasing sequence of compacts. Suppose that $\bigcap A_n=\emptyset$. Take $U_n=A_n^c$, then
$$
\bigcup U_n=\bigcup A_n^c=\left(\bigcap A_n\right)^c=A_1 .
$$
Here, I'm thinking of $A_1$ as the main metric space. Since $\left\{U_n\right\}$ ... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
323 | Pugh_exercise_3_1 | Pugh | import Mathlib
open Filter Real Function
open scoped Topology | Assume that $f \colon \mathbb{R} \rightarrow \mathbb{R}$ satisfies $|f(t)-f(x)| \leq|t-x|^{2}$ for all $t, x$. Prove that $f$ is constant. | theorem Pugh_exercise_3_1 {f : ℝ → ℝ}
(hf : ∀ x y, |f x - f y| ≤ |x - y| ^ 2) :
∃ c, f = λ _x => c := by
sorry | We have $|f(t)-f(x)| \leq|t-x|^2, \forall t, x \in \mathbb{R}$. Fix $x \in \mathbb{R}$ and let $t \neq x$. Then
$$
\left|\frac{f(t)-f(x)}{t-x}\right| \leq|t-x| \text {, hence } \lim _{t \rightarrow x}\left|\frac{f(t)-f(x)}{t-x}\right|=0 \text {, }
$$
so $f$ is differentiable in $\mathbb{R}$ and $f^{\prime}=0$. This imp... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
324 | Pugh_exercise_3_63a | Pugh | import Mathlib
open Filter Real Function
open scoped Topology | Prove that $\sum 1/k(\log(k))^p$ converges when $p > 1$. | theorem Pugh_exercise_3_63a (p : ℝ) (hp : p > 1) :
Summable (fun (k : ℕ) => (1 : ℝ) / ((k + 2) * (Real.log (k + 2)) ^ p)) := by
sorry | Using the integral test, for a set $a$, we see
$$
\lim _{b \rightarrow \infty} \int_a^b \frac{1}{x \log (x)^c} d x=\lim _{b \rightarrow \infty}\left(\frac{\log (b)^{1-c}}{1-c}-\frac{\log (a)^{1-c}}{1-c}\right)
$$
which goes to infinity if $c \leq 1$ and converges if $c>1$. Thus,
$$
\sum_{n=2}^{\infty} \frac{1}{n \log (... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
325 | Pugh_exercise_4_15a | Pugh | import Mathlib
open Filter Real Function Set
open scoped Topology | A continuous, strictly increasing function $\mu \colon (0, \infty) \rightarrow (0, \infty)$ is a modulus of continuity if $\mu(s) \rightarrow 0$ as $s \rightarrow 0$. A function $f \colon [a, b] \rightarrow \mathbb{R}$ has modulus of continuity $\mu$ if $|f(s) - f(t)| \leq \mu(|s - t|)$ for all $s, t \in [a, b]$. Prove... | theorem Pugh_exercise_4_15a (a b : ℝ) (hab : a ≤ b) (f : Icc a b → ℝ) :
UniformContinuous f ↔ ∃ (μ : ℝ → ℝ), (∀ x : ℝ, x > 0 → μ x > 0) ∧
ContinuousOn μ (Ioi 0) ∧ StrictMonoOn μ (Ioi 0) ∧ Tendsto μ (𝓝[>] 0) (𝓝 0) ∧
(∀ s t : Icc a b, |f s - f t| ≤ μ (|s - t|)) := by
sorry | Suppose there exists a modulus of continuity $w$ for $f$, then fix $\varepsilon>0$, since $\lim _{s \rightarrow 0} w(s)=0$, there exists $\delta>0$ such that for any $|s|<\delta$, we have $w(s)<\varepsilon$, then we have for any $x, z \in X$ such that $d_X(x, z)<\delta$, we have $d_Y(f(x), f(z)) \leq w\left(d_X(x, z)\r... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
326 | Putnam_exercise_1998_b6 | Putnam | import Mathlib
open scoped BigOperators | Prove that, for any integers $a, b, c$, there exists a positive integer $n$ such that $\sqrt{n^3+a n^2+b n+c}$ is not an integer. | theorem Putnam_exercise_1998_b6 (a b c : ℤ) :
∃ n : ℤ, n > 0 ∧ ¬ ∃ m : ℤ, Real.sqrt (n^3 + a*n^2 + b*n + c) = m := by
sorry | We prove more generally that for any polynomial $P(z)$ with integer
coefficients which is not a perfect square, there exists a positive
integer $n$ such that $P(n)$ is not a perfect square. Of course it
suffices to assume $P(z)$ has no repeated factors, which is to say $P(z)$
and its derivative $P'(z)$ are relatively p... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
327 | Putnam_exercise_1999_b4 | Putnam | import Mathlib
set_option maxHeartbeats 800000
open scoped BigOperators
open Filter Topology Set | Let $f$ be a real function with a continuous third derivative such that $f(x), f^{\prime}(x), f^{\prime \prime}(x), f^{\prime \prime \prime}(x)$ are positive for all $x$. Suppose that $f^{\prime \prime \prime}(x) \leq f(x)$ for all $x$. Show that $f^{\prime}(x)<2 f(x)$ for all $x$. | theorem Putnam_exercise_1999_b4 (f : ℝ → ℝ) (hf: ContDiff ℝ 3 f)
(hf1 : ∀ n ≤ 3, ∀ x : ℝ, iteratedDeriv n f x > 0)
(hf2 : ∀ x : ℝ, iteratedDeriv 3 f x ≤ f x) :
∀ x : ℝ, deriv f x < 2 * f x := by
sorry | \setcounter{equation}{0}
We make repeated use of the following fact: if $f$ is a differentiable function on all of
$\mathbb{R}$, $\lim_{x \to -\infty} f(x) \geq 0$, and $f'(x) > 0$ for all $x \in \mathbb{R}$, then
$f(x) > 0$ for all $x \in \mathbb{R}$. (Proof: if $f(y) < 0$ for some $x$, then $f(x)< f(y)$ for all
$x<y$... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
328 | Putnam_exercise_2001_a5 | Putnam | import Mathlib
open scoped BigOperators | Prove that there are unique positive integers $a, n$ such that $a^{n+1}-(a+1)^n=2001$. | theorem Putnam_exercise_2001_a5 :
∃! p : ℕ × ℕ, p.1 > 0 ∧ p.2 > 0 ∧ p.1^(p.2 + 1) - (p.1 + 1)^(p.2) = 2001 := by
sorry | Suppose $a^{n+1} - (a+1)^n = 2001$.
Notice that $a^{n+1} + [(a+1)^n - 1]$ is a multiple of $a$; thus
$a$ divides $2002 = 2 \times 7 \times 11 \times 13$.
Since $2001$ is divisible by 3, we must have $a \equiv 1 \pmod{3}$,
otherwise one of $a^{n+1}$ and $(a+1)^n$ is a multiple of 3 and the
other is not, so their differ... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
329 | Putnam_exercise_2014_a5 | Putnam | import Mathlib
open scoped BigOperators | Let $P_n(x)=1+2 x+3 x^2+\cdots+n x^{n-1} .$ Prove that the polynomials $P_j(x)$ and $P_k(x)$ are relatively prime for all positive integers $j$ and $k$ with $j \neq k$. | theorem Putnam_exercise_2014_a5
(P : ℕ → Polynomial ℚ)
(hP : ∀ n, P n = ∑ i : Fin n, (↑(i : ℕ) + 1) * Polynomial.X ^ (i : ℕ)) :
∀ {j k : ℕ}, 0 < j → 0 < k → j ≠ k → IsCoprime (P j) (P k) := by
sorry | Suppose to the contrary that there exist positive integers $i \neq j$ and a complex number $z$ such that $P_i(z) = P_j(z) = 0$. Note that $z$ cannot be a nonnegative real number or else $P_i(z), P_j(z) > 0$; we may put $w = z^{-1} \neq 0,1$. For $n \in \{i+1,j+1\}$ we compute that
\[
w^n = n w - n + 1,
\qquad \overline... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
330 | Putnam_exercise_2018_a5 | Putnam | import Mathlib
open scoped BigOperators | Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be an infinitely differentiable function satisfying $f(0)=0, f(1)=1$, and $f(x) \geq 0$ for all $x \in$ $\mathbb{R}$. Show that there exist a positive integer $n$ and a real number $x$ such that $f^{(n)}(x)<0$. | theorem Putnam_exercise_2018_a5 (f : ℝ → ℝ) (hf : ContDiff ℝ ((⊤ : ℕ∞) : WithTop ℕ∞) f)
(hf0 : f 0 = 0) (hf1 : f 1 = 1) (hf2 : ∀ x, f x ≥ 0) :
∃ (n : ℕ+) (x : ℝ), iteratedDeriv n f x < 0 := by
sorry | Call a function $f\colon \mathbb{R} \to \mathbb{R}$ \textit{ultraconvex} if $f$ is infinitely differentiable and $f^{(n)}(x) \geq 0$ for all $n \geq 0$ and all $x \in \mathbb{R}$, where $f^{(0)}(x) = f(x)$;
note that if $f$ is ultraconvex, then so is $f'$.
Define the set
\[
S = \{ f :\thinspace \mathbb{R} \to \mathbb{R... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
331 | Putnam_exercise_2018_b4 | Putnam | import Mathlib
open scoped BigOperators
open Real Nat Function | Given a real number $a$, we define a sequence by $x_{0}=1$, $x_{1}=x_{2}=a$, and $x_{n+1}=2 x_{n} x_{n-1}-x_{n-2}$ for $n \geq 2$. Prove that if $x_{n}=0$ for some $n$, then the sequence is periodic. | theorem Putnam_exercise_2018_b4 (a : ℝ) (x : ℕ → ℝ) (hx0 : x 0 = 1)
(hx1 : x 1 = a) (hx2 : x 2 = a)
(hxn : ∀ n : ℕ, n ≥ 2 → x (n+1) = 2*(x n)*(x (n-1)) - x (n-2))
(h : ∃ n, x n = 0) :
∃ c ≠ 0, Periodic x c := by
sorry | We first rule out the case $|a|>1$. In this case, we prove that $|x_{n+1}| \geq |x_n|$ for all $n$, meaning that we cannot have $x_n = 0$. We proceed by induction; the claim is true for $n=0,1$ by hypothesis. To prove the claim for $n \geq 2$, write
\begin{align*}
|x_{n+1}| &= |2x_nx_{n-1}-x_{n-2}| \\
&\geq 2|x_n||x_{... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
332 | Rudin_exercise_1_1b | Rudin | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators | If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $rx$ is irrational. | theorem Rudin_exercise_1_1b
(x : ℝ)
(y : ℚ)
(h : y ≠ 0)
: ( Irrational x ) -> Irrational ( x * y ) := by
sorry | If $r x$ were rational, then $x=\frac{r x}{r}$ would also be rational. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
333 | Rudin_exercise_1_4 | Rudin | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators | Let $E$ be a nonempty subset of a total ordered set; suppose $\alpha$ is a lower bound of $E$ and $\beta$ is an upper bound of $E$. Prove that $\alpha \leq \beta$. | theorem Rudin_exercise_1_4
(α : Type*) [LinearOrder α]
(s : Set α)
(x y : α)
(h₀ : Set.Nonempty s)
(h₁ : x ∈ lowerBounds s)
(h₂ : y ∈ upperBounds s)
: x ≤ y := by
sorry | Since $E$ is nonempty, there exists $x \in E$. Then by definition of lower and upper bounds we have $\alpha \leq x \leq \beta$, and hence by property $i i$ in the definition of an ordering, we have $\alpha<\beta$ unless $\alpha=x=\beta$. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
334 | Rudin_exercise_1_8 | Rudin | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators | Prove that no order can be defined in the complex field that turns it into an ordered field. | theorem Rudin_exercise_1_8 :
¬ (∃ inst : LinearOrder ℂ, by
let _ : LinearOrder ℂ := inst
exact IsStrictOrderedRing ℂ) := by
sorry | By Part (a) of Proposition $1.18$, either $i$ or $-i$ must be positive. Hence $-1=i^2=(-i)^2$ must be positive. But then $1=(-1)^2$, must also be positive, and this contradicts Part $(a)$ of Proposition 1.18, since 1 and $-1$ cannot both be positive. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
335 | Rudin_exercise_1_12 | Rudin | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators | If $z_1, \ldots, z_n$ are complex, prove that $|z_1 + z_2 + \ldots + z_n| \leq |z_1| + |z_2| + \cdots + |z_n|$. | theorem Rudin_exercise_1_12 (n : ℕ) (f : ℕ → ℂ) :
‖∑ i ∈ range n, f i‖ ≤ ∑ i ∈ range n, ‖f i‖ := by
sorry | We can apply the case $n=2$ and induction on $n$ to get
$$
\begin{aligned}
\left|z_1+z_2+\cdots z_n\right| &=\left|\left(z_1+z_2+\cdots+z_{n-1}\right)+z_n\right| \\
& \leq\left|z_1+z_2+\cdots+z_{n-1}\right|+\left|z_n\right| \\
& \leq\left|z_1\right|+\left|z_2\right|+\cdots+\left|z_{n-1}\right|+\left|z_n\right|
\end{ali... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
336 | Rudin_exercise_1_17 | Rudin | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators | Prove that $|\mathbf{x}+\mathbf{y}|^{2}+|\mathbf{x}-\mathbf{y}|^{2}=2|\mathbf{x}|^{2}+2|\mathbf{y}|^{2}$ if $\mathbf{x} \in R^{k}$ and $\mathbf{y} \in R^{k}$. | theorem Rudin_exercise_1_17
(n : ℕ)
(x y : EuclideanSpace ℝ (Fin n)) -- R^n
: ‖x + y‖^2 + ‖x - y‖^2 = 2*‖x‖^2 + 2*‖y‖^2 := by
sorry | The proof is a routine computation, using the relation
$$
|x \pm y|^2=(x \pm y) \cdot(x \pm y)=|x|^2 \pm 2 x \cdot y+|y|^2 .
$$
If $\mathrm{x}$ and $\mathrm{y}$ are the sides of a parallelogram, then $\mathrm{x}+\mathrm{y}$ and $\mathbf{x}-\mathrm{y}$ are its diagonals. Hence this result says that the sum of the square... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
337 | Rudin_exercise_1_18b | Rudin | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators | If $k = 1$ and $\mathbf{x} \in R^{k}$, $\mathbf{x} \neq 0$, prove that there does not exist $\mathbf{y} \in R^{k}$ such that $\mathbf{y} \neq 0$ but $\mathbf{x} \cdot \mathbf{y}=0$ | theorem Rudin_exercise_1_18b
: ∀ (x : ℝ), x ≠ 0 → ¬ ∃ (y : ℝ), y ≠ 0 ∧ x * y = 0 := by
sorry | Not true when $k=1$, since the product of two nonzero real numbers is nonzero. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
338 | Rudin_exercise_2_19a | Rudin | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators | If $A$ and $B$ are disjoint closed sets in some metric space $X$, prove that they are separated. | theorem Rudin_exercise_2_19a {X : Type*} [MetricSpace X]
(A B : Set X) (hA : IsClosed A) (hB : IsClosed B) (hAB : Disjoint A B) :
A ∩ closure B = ∅ ∧ closure A ∩ B = ∅ := by
sorry | We are given that $A \cap B=\varnothing$. Since $A$ and $B$ are closed, this means $A \cap \bar{B}=\varnothing=\bar{A} \cap B$, which says that $A$ and $B$ are separated. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
339 | Rudin_exercise_2_25 | Rudin | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators | Prove that every compact metric space $K$ has a countable base. | theorem Rudin_exercise_2_25 {K : Type*} [MetricSpace K] [CompactSpace K] :
∃ (B : Set (Set K)), Set.Countable B ∧ IsTopologicalBasis B := by
sorry | $K$ can be covered by a finite union of neighborhoods of radius $1 / n$, and this shows that this implies that $K$ is separable.
It is not entirely obvious that a metric space with a countable base is separable. To prove this, let $\left\{V_n\right\}_{n=1}^{\infty}$ be a countable base, and let $x_n \in V_n$. The poin... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
340 | Rudin_exercise_2_27b | Rudin | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators | Suppose $E\subset\mathbb{R}^k$ is uncountable, and let $P$ be the set of condensation points of $E$. Prove that at most countably many points of $E$ are not in $P$. | theorem Rudin_exercise_2_27b (k : ℕ) (E P : Set (EuclideanSpace ℝ (Fin k)))
(hE : ¬ Set.Countable E)
(hP : P = {x | ∀ U ∈ 𝓝 x, ¬ Set.Countable (U ∩ E)}) :
Set.Countable (E \ P) := by
sorry | If $x \in W^c$, and $O$ is any neighborhood of $x$, then $x \in V_n \subseteq O$ for some n. Since $x \notin W, V_n \cap E$ is uncountable. Hence $O$ contains uncountably many points of $E$, and so $x$ is a condensation point of $E$. Thus $x \in P$, i.e., $W^c \subseteq P$.
Conversely if $x \in W$, then $x \in V_n$ for... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
341 | Rudin_exercise_2_29 | Rudin | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset Set
open scoped BigOperators | Prove that every open set in $\mathbb{R}$ is the union of an at most countable collection of disjoint segments. | theorem Rudin_exercise_2_29 (U : Set ℝ) (hU : IsOpen U) :
∃ (f : ℕ → Set ℝ),
(∀ n, ∃ a b : EReal, f n = {x : ℝ | a < (x : EReal) ∧ (x : EReal) < b}) ∧
(∀ n, f n ⊆ U) ∧
(∀ n m, n ≠ m → f n ∩ f m = ∅) ∧
U = ⋃ n, f n := by
sorry | Let $O$ be open. For each pair of points $x \in O, y \in O$, we define an equivalence relation $x \sim y$ by saying $x \sim y$ if and only if $[\min (x, y), \max (x, y)] \subset$ 0 . This is an equivalence relation, since $x \sim x([x, x] \subset O$ if $x \in O)$; if $x \sim y$, then $y \sim x$ (since $\min (x, y)=\min... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
342 | Rudin_exercise_3_2a | Rudin | import Mathlib
open Topology Filter Real TopologicalSpace Finset
open scoped BigOperators | Prove that $\lim_{n \rightarrow \infty}\sqrt{n^2 + n} - n = 1/2$ for the real function $f(n) = \sqrt{n^2+n} - n$. | theorem Rudin_exercise_3_2a
: Tendsto (λ (n : ℝ) => (sqrt (n^2 + n) - n)) atTop (𝓝 (1/2)) := by
sorry | Multiplying and dividing by $\sqrt{n^2+n}+n$ yields
$$
\sqrt{n^2+n}-n=\frac{n}{\sqrt{n^2+n}+n}=\frac{1}{\sqrt{1+\frac{1}{n}}+1} .
$$
It follows that the limit is $\frac{1}{2}$. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
343 | Rudin_exercise_3_5 | Rudin | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators | local notation "limsup" => fun u : ℕ → EReal => Filter.limsup u Filter.atTop | For any two real sequences $\left\{a_{n}\right\},\left\{b_{n}\right\}$, prove that $\limsup _{n \rightarrow \infty}\left(a_{n}+b_{n}\right) \leq \limsup _{n \rightarrow \infty} a_{n}+\limsup _{n \rightarrow \infty} b_{n},$ provided the sum on the right is not of the form $\infty-\infty$. | theorem Rudin_exercise_3_5
(a b : ℕ → ℝ)
(h1 : limsup (fun n => (a n : EReal)) ≠ ⊥ ∨ limsup (fun n => (b n : EReal)) ≠ ⊤)
(h2 : limsup (fun n => (a n : EReal)) ≠ ⊤ ∨ limsup (fun n => (b n : EReal)) ≠ ⊥) :
limsup (fun n => ((a n + b n : ℝ) : EReal)) ≤
limsup (fun n => (a n : EReal)) + limsup (fun n... | Since the case when $\limsup _{n \rightarrow \infty} a_n=+\infty$ and $\limsup _{n \rightarrow \infty} b_n=-\infty$ has been excluded from consideration, we note that the inequality is obvious if $\limsup _{n \rightarrow \infty} a_n=+\infty$. Hence we shall assume that $\left\{a_n\right\}$ is bounded above.
Let $\left... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip |
344 | Rudin_exercise_3_7 | Rudin | import Mathlib
open Topology Filter Real TopologicalSpace Finset
open scoped BigOperators | Prove that the convergence of $\Sigma a_{n}$ implies the convergence of $\sum \frac{\sqrt{a_{n}}}{n}$ if $a_n\geq 0$. | theorem Rudin_exercise_3_7
(a : ℕ → ℝ)
(hnneg : ∀ n, a n ≥ 0)
(h : ∃ y, (Tendsto (λ n => (∑ i ∈ (range n), a i)) atTop (𝓝 y))) :
∃ y, Tendsto (λ n => (∑ i ∈ (range n), sqrt (a i) / (i + 1))) atTop (𝓝 y) := by
sorry | Since $\left(\sqrt{a_n}-\frac{1}{n}\right)^2 \geq 0$, it follows that
$$
\frac{\sqrt{a_n}}{n} \leq \frac{1}{2}\left(a_n^2+\frac{1}{n^2}\right) .
$$
Now $\Sigma a_n^2$ converges by comparison with $\Sigma a_n$ (since $\Sigma a_n$ converges, we have $a_n<1$ for large $n$, and hence $\left.a_n^2<a_n\right)$. Since $\Sigma... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
345 | Rudin_exercise_3_13 | Rudin | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators | Prove that the Cauchy product of two absolutely convergent series converges absolutely. | theorem Rudin_exercise_3_13
(a b : ℕ → ℝ)
(ha : ∃ y, Tendsto (fun n => ∑ i ∈ range n, |a i|) atTop (𝓝 y))
(hb : ∃ y, Tendsto (fun n => ∑ i ∈ range n, |b i|) atTop (𝓝 y)) :
∃ y,
Tendsto
(fun n =>
∑ i ∈ range n,
|∑ j ∈ range (i + 1), a j * b (i - j)|)
atTop (𝓝 y) := b... | Since both the hypothesis and conclusion refer to absolute convergence, we may assume both series consist of nonnegative terms. We let $S_n=\sum_{k=0}^n a_n, T_n=\sum_{k=0}^n b_n$, and $U_n=\sum_{k=0}^n \sum_{l=0}^k a_l b_{k-l}$. We need to show that $U_n$ remains bounded, given that $S_n$ and $T_n$ are bounded. To do ... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
346 | Rudin_exercise_3_21 | Rudin | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators | If $\left\{E_{n}\right\}$ is a sequence of closed nonempty and bounded sets in a complete metric space $X$, if $E_{n} \supset E_{n+1}$, and if $\lim _{n \rightarrow \infty} \operatorname{diam} E_{n}=0,$ then $\bigcap_{1}^{\infty} E_{n}$ consists of exactly one point. | theorem Rudin_exercise_3_21
{X : Type*} [MetricSpace X] [CompleteSpace X]
(E : ℕ → Set X)
(hEcl : ∀ n, IsClosed (E n))
(hEne : ∀ n, (E n).Nonempty)
(hEbdd : ∀ n, Bornology.IsBounded (E n))
(hE : ∀ n, E n ⊇ E (n + 1))
(hE' : Tendsto (λ n => Metric.diam (E n)) atTop (𝓝 0)) :
∃ a, Set.iInter E = {a} := by... | Choose $x_n \in E_n$. (We use the axiom of choice here.) The sequence $\left\{x_n\right\}$ is a Cauchy sequence, since the diameter of $E_n$ tends to zero as $n$ tends to infinity and $E_n$ contains $E_{n+1}$. Since the metric space $X$ is complete, the sequence $x_n$ converges to a point $x$, which must belong to $E_n... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
347 | Rudin_exercise_4_1a | Rudin | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators | Suppose $f$ is a real function defined on $\mathbb{R}$ which satisfies $\lim_{h \rightarrow 0} f(x + h) - f(x - h) = 0$ for every $x \in \mathbb{R}$. Show that $f$ does not need to be continuous. | theorem Rudin_exercise_4_1a
: ∃ (f : ℝ → ℝ), (∀ (x : ℝ), Tendsto (λ y => f (x + y) - f (x - y)) (𝓝 0) (𝓝 0)) ∧ ¬ Continuous f := by
sorry | $$
f(x)= \begin{cases}1 & \text { if } x \text { is an integer } \\ 0 & \text { if } x \text { is not an integer. }\end{cases}
$$
(If $x$ is an integer, then $f(x+h)-f(x-h) \equiv 0$ for all $h$; while if $x$ is not an integer, $f(x+h)-f(x-h)=0$ for $|h|<\min (x-[x], 1+[x]-x)$. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
348 | Rudin_exercise_4_3 | Rudin | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators | Let $f$ be a continuous real function on a metric space $X$. Let $Z(f)$ (the zero set of $f$ ) be the set of all $p \in X$ at which $f(p)=0$. Prove that $Z(f)$ is closed. | theorem Rudin_exercise_4_3
{α : Type} [MetricSpace α]
(f : α → ℝ) (h : Continuous f) (z : Set α) (g : z = f⁻¹' {0})
: IsClosed z := by
sorry | $Z(f)=f^{-1}(\{0\})$, which is the inverse image of a closed set. Hence $Z(f)$ is closed. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
349 | Rudin_exercise_4_4b | Rudin | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators | Let $f$ and $g$ be continuous mappings of a metric space $X$ into a metric space $Y$, and let $P$ be a dense subset of $X$. Prove that if $g(p) = f(p)$ for all $p \in P$ then $g(p) = f(p)$ for all $p \in X$. | theorem Rudin_exercise_4_4b
{α : Type} [MetricSpace α]
{β : Type} [MetricSpace β]
(f g : α → β)
(s : Set α)
(h₁ : Continuous f)
(h₂ : Continuous g)
(h₃ : Dense s)
(h₄ : ∀ x ∈ s, f x = g x)
: f = g := by
sorry | The function $\varphi: X \rightarrow R^1$ given by
$$
\varphi(p)=d_Y(f(p), g(p))
$$
is continuous, since
$$
\left|d_Y(f(p), g(p))-d_Y(f(q), g(q))\right| \leq d_Y(f(p), f(q))+d_Y(g(p), g(q))
$$
(This inequality follows from the triangle inequality, since
$$
d_Y(f(p), g(p)) \leq d_Y(f(p), f(q))+d_Y(f(q), g(q))+d_Y(g(q), ... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
350 | Rudin_exercise_4_5b | Rudin | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators | Show that there exist a set $E \subset \mathbb{R}$ and a real continuous function $f$ defined on $E$, such that there does not exist a continuous real function $g$ on $\mathbb{R}$ such that $g(x)=f(x)$ for all $x \in E$. | theorem Rudin_exercise_4_5b
: ∃ (E : Set ℝ) (f : ℝ → ℝ), (ContinuousOn f E) ∧
(¬ ∃ (g : ℝ → ℝ), Continuous g ∧ ∀ x ∈ E, f x = g x) := by
sorry | Let $E:=(0,1)$, and define $f(x):=1 / x$ for all $x \in E$. If $f$ has a continuous extension $g$ to $\mathbb{R}$, then $g$ is continuous on $[-1,1]$, and therefore bounded by the intermediate value theorem. However, $f$ is not bounded in any neighborhood of $x=0$, so therefore $g$ is not bounded either, a contradictio... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
351 | Rudin_exercise_4_8a | Rudin | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators | Let $f$ be a real uniformly continuous function on the bounded set $E$ in $R^{1}$. Prove that $f$ is bounded on $E$. | theorem Rudin_exercise_4_8a
(E : Set ℝ) (f : ℝ → ℝ) (hf : UniformContinuousOn f E)
(hE : Bornology.IsBounded E) : Bornology.IsBounded (Set.image f E) := by
sorry | Let $a=\inf E$ and $b=\sup E$, and let $\delta>0$ be such that $|f(x)-f(y)|<1$ if $x, y \in E$ and $|x-y|<\delta$. Now choose a positive integer $N$ larger than $(b-a) / \delta$, and consider the $N$ intervals $I_k=\left[a+\frac{k-1}{b-a}, a+\frac{k}{b-a}\right], k=1,2, \ldots, N$. For each $k$ such that $I_k \cap E \n... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
352 | Rudin_exercise_4_11a | Rudin | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators | Suppose $f$ is a uniformly continuous mapping of a metric space $X$ into a metric space $Y$ and prove that $\left\{f\left(x_{n}\right)\right\}$ is a Cauchy sequence in $Y$ for every Cauchy sequence $\{x_n\}$ in $X$. | theorem Rudin_exercise_4_11a
{X : Type*} [MetricSpace X]
{Y : Type*} [MetricSpace Y]
(f : X → Y) (hf : UniformContinuous f)
(x : ℕ → X) (hx : CauchySeq x) :
CauchySeq (λ n => f (x n)) := by
sorry | Suppose $\left\{x_n\right\}$ is a Cauchy sequence in $X$. Let $\varepsilon>0$ be given. Let $\delta>0$ be such that $d_Y(f(x), f(u))<\varepsilon$ if $d_X(x, u)<\delta$. Then choose $N$ so that $d_X\left(x_n, x_m\right)<\delta$ if $n, m>N$. Obviously $d_Y\left(f\left(x_n\right), f\left(x_m\right)\right)<\varepsilon$ if ... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
353 | Rudin_exercise_4_15 | Rudin | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset Set
open scoped BigOperators | Prove that every continuous open mapping of $R^{1}$ into $R^{1}$ is monotonic. | theorem Rudin_exercise_4_15 {f : ℝ → ℝ}
(hf : Continuous f) (hof : IsOpenMap f) :
Monotone f ∨ Antitone f := by
sorry | Suppose $f$ is continuous and not monotonic, say there exist points $a<b<c$ with $f(a)<f(b)$, and $f(c)<f(b)$. Then the maximum value of $f$ on the closed interval $[a, c]$ is assumed at a point $u$ in the open interval $(a, c)$. If there is also a point $v$ in the open interval $(a, c)$ where $f$ assumes its minimum v... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
354 | Rudin_exercise_4_21a | Rudin | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset Metric
open scoped BigOperators | Suppose $K$ and $F$ are disjoint sets in a metric space $X, K$ is compact, $F$ is closed. Prove that there exists $\delta>0$ such that $d(p, q)>\delta$ if $p \in K, q \in F$. | theorem Rudin_exercise_4_21a {X : Type*} [MetricSpace X]
(K F : Set X) (hK : IsCompact K) (hF : IsClosed F) (hKF : Disjoint K F) :
∃ (δ : ℝ), δ > 0 ∧ ∀ (p q : X), p ∈ K → q ∈ F → dist p q > δ := by
sorry | Following the hint, we observe that $\rho_F(x)$ must attain its minimum value on $K$, i.e., there is some point $r \in K$ such that
$$
\rho_F(r)=\min _{q \in K} \rho_F(q) .
$$
Since $F$ is closed and $r \notin F$, it follows from Exercise $4.20$ that $\rho_F(r)>0$. Let $\delta$ be any positive number smaller than $\rho... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
355 | Rudin_exercise_5_1 | Rudin | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators | Let $f$ be defined for all real $x$, and suppose that $|f(x)-f(y)| \leq (x-y)^{2}$ for all real $x$ and $y$. Prove that $f$ is constant. | theorem Rudin_exercise_5_1
{f : ℝ → ℝ} (hf : ∀ x y : ℝ, |(f x - f y)| ≤ (x - y) ^ 2) :
∃ c, f = λ _x => c := by
sorry | Dividing by $x-y$, and letting $x \rightarrow y$, we find that $f^{\prime}(y)=0$ for all $y$. Hence $f$ is constant. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
356 | Rudin_exercise_5_3 | Rudin | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators | Suppose $g$ is a real function on $R^{1}$, with bounded derivative (say $\left|g^{\prime}\right| \leq M$ ). Fix $\varepsilon>0$, and define $f(x)=x+\varepsilon g(x)$. Prove that $f$ is one-to-one if $\varepsilon$ is small enough. | theorem Rudin_exercise_5_3
{g : ℝ → ℝ} (hgDiff : Differentiable ℝ g)
(hg' : ∃ M : ℝ, ∀ x : ℝ, |deriv g x| ≤ M) :
∃ N > 0, ∀ ε > 0, ε < N → Function.Injective (λ x : ℝ => x + ε * g x) := by
sorry | If $0<\varepsilon<\frac{1}{M}$, we certainly have
$$
f^{\prime}(x) \geq 1-\varepsilon M>0,
$$
and this implies that $f(x)$ is one-to-one, by the preceding problem. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
357 | Rudin_exercise_5_5 | Rudin | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset Set
open scoped BigOperators | Suppose $f$ is defined and differentiable for every $x>0$, and $f^{\prime}(x) \rightarrow 0$ as $x \rightarrow+\infty$. Put $g(x)=f(x+1)-f(x)$. Prove that $g(x) \rightarrow 0$ as $x \rightarrow+\infty$. | theorem Rudin_exercise_5_5
{f : ℝ → ℝ}
(hfd : DifferentiableOn ℝ f (Ioi 0))
(hf : Tendsto (deriv f) atTop (𝓝 0)) :
Tendsto (λ x => f (x + 1) - f x) atTop (𝓝 0) := by
sorry | Let $\varepsilon>0$. Choose $x_0$ such that $\left|f^{\prime}(x)\right|<\varepsilon$ if $x>x_0$. Then for any $x \geq x_0$ there exists $x_1 \in(x, x+1)$ such that
$$
f(x+1)-f(x)=f^{\prime}\left(x_1\right) .
$$
Since $\left|f^{\prime}\left(x_1\right)\right|<\varepsilon$, it follows that $|f(x+1)-f(x)|<\varepsilon$, as ... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
358 | Rudin_exercise_5_7 | Rudin | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators | Suppose $f^{\prime}(x), g^{\prime}(x)$ exist, $g^{\prime}(x) \neq 0$, and $f(x)=g(x)=0$. Prove that $\lim _{t \rightarrow x} \frac{f(t)}{g(t)}=\frac{f^{\prime}(x)}{g^{\prime}(x)}.$ | theorem Rudin_exercise_5_7
{f g : ℝ → ℝ} {x : ℝ}
(hf' : DifferentiableAt ℝ f x)
(hg' : DifferentiableAt ℝ g x)
(hg'_ne_0 : deriv g x ≠ 0)
(f0 : f x = 0) (g0 : g x = 0) :
Tendsto (λ t => f t / g t) (𝓝[≠] x) (𝓝 (deriv f x / deriv g x)) := by
sorry | Since $f(x)=g(x)=0$, we have
$$
\begin{aligned}
\lim _{t \rightarrow x} \frac{f(t)}{g(t)} &=\lim _{t \rightarrow x} \frac{\frac{f(t)-f(x)}{t-x}}{\frac{g(t)-g(x)}{t-x}} \\
&=\frac{\lim _{t \rightarrow x} \frac{f(t)-f(x)}{t-x}}{\lim _{t \rightarrow x} \frac{g(t)-g(x)}{t-x}} \\
&=\frac{f^{\prime}(x)}{g^{\prime}(x)}
\end{a... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
359 | Rudin_exercise_5_17 | Rudin | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset Set
open scoped BigOperators | Suppose $f$ is a real, three times differentiable function on $[-1,1]$, such that $f(-1)=0, \quad f(0)=0, \quad f(1)=1, \quad f^{\prime}(0)=0 .$ Prove that $f^{(3)}(x) \geq 3$ for some $x \in(-1,1)$. | theorem Rudin_exercise_5_17
{f : ℝ → ℝ}
(hf' : DifferentiableOn ℝ f (Icc (-1) 1))
(hf'' : DifferentiableOn ℝ (deriv f) (Icc (-1) 1))
(hf''' : DifferentiableOn ℝ (deriv (deriv f)) (Icc (-1) 1))
(hf0 : f (-1) = 0)
(hf1 : f 0 = 0)
(hf2 : f 1 = 1)
(hf3 : deriv f 0 = 0) :
∃ x, x ∈ Ioo (-1 : ℝ) 1 ∧ deriv (d... | Following the hint, we observe that Theorem $5.15$ (Taylor's formula with remainder) implies that
$$
\begin{aligned}
f(1) &=f(0)+f^{\prime}(0)+\frac{1}{2} f^{\prime \prime}(0)+\frac{1}{6} f^{(3)}(s) \\
f(-1) &=f(0)-f^{\prime}(0)+\frac{1}{2} f^{\prime \prime}(0)-\frac{1}{6} f^{(3)}(t)
\end{aligned}
$$
for some $s \in(0,... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
360 | Shakarchi_exercise_1_13b | Shakarchi | import Mathlib
open Complex Filter Function Metric Finset
open scoped BigOperators Topology | Suppose that $f$ is holomorphic in an open set $\Omega$. Prove that if $\text{Im}(f)$ is constant, then $f$ is constant. | theorem Shakarchi_exercise_1_13b {f : ℂ → ℂ} {Ω : Set ℂ}
(hΩ : IsOpen Ω) (hConn : IsPreconnected Ω)
(hf : DifferentiableOn ℂ f Ω)
(hc : ∃ c : ℝ, ∀ z ∈ Ω, (f z).im = c) :
∀ a b : Ω, f a = f b := by
sorry | Let $f(z)=f(x, y)=u(x, y)+i v(x, y)$, where $z=x+i y$.
Since $\operatorname{Im}(f)=$ constant,
$$
\frac{\partial v}{\partial x}=0, \frac{\partial v}{\partial y}=0 .
$$
By the Cauchy-Riemann equations,
$$
\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}=0 .
$$
Thus in $\Omega$,
$$
f^{\prime}(z)=\frac{\partial... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
361 | Shakarchi_exercise_1_19a | Shakarchi | import Mathlib
open Complex Filter Function Metric Finset
open scoped BigOperators Topology | Prove that the power series $\sum nz^n$ does not converge on any point of the unit circle. | theorem Shakarchi_exercise_1_19a (z : ℂ) (hz : ‖z‖ = 1) (s : ℕ → ℂ)
(h : s = (λ n => ∑ i ∈ (range n), i * z ^ i)) :
¬ ∃ y, Tendsto s atTop (𝓝 y) := by
sorry | For $z \in S:=\{z \in \mathbb{C}:|z|=1\}$ it also holds $z^n \in S$ for all $n \in \mathbb{N}$ (since in this case $\left.\left|z^n\right|=|z|^n=1^n=1\right)$
Thus, the sequence $\left(a_n\right)_{n \in \mathbb{N}}$ with $a_n=n z^n$ does not converge to zero which is necessary for the corresponding sum $\sum_{n \in \ma... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
362 | Shakarchi_exercise_1_19c | Shakarchi | import Mathlib
open Complex Filter Function Metric Finset
open scoped BigOperators Topology | Prove that the power series $\sum z^n/n$ converges at every point of the unit circle except $z = 1$. | theorem Shakarchi_exercise_1_19c (z : ℂ) (hz : ‖z‖ = 1) (s : ℕ → ℂ)
(h : s = (λ n => ∑ i ∈ (range n), z ^ (i + 1) / (i + 1))) :
(∃ z₀, Tendsto s atTop (𝓝 z₀)) ↔ z ≠ 1 := by
sorry | If $z=1$ then $\sum z^n / n=\sum 1 / n$ is divergent (harmonic series). If $|z|=1$ and $z \neq 1$, write $z=e^{2 \pi i t}$ with $t \in(0,1)$ and apply Dirichlet's test: if $\left\{a_n\right\}$ is a sequence of real numbers and $\left\{b_n\right\}$ a sequence of complex numbers satisfying
- $a_{n+1} \leq a_n$
- $\lim _{... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
363 | Shakarchi_exercise_2_2 | Shakarchi | import Mathlib
open Complex Filter Function Metric Finset MeasureTheory intervalIntegral
open scoped BigOperators Topology | Show that $\int_{0}^{\infty} \frac{\sin x}{x} d x=\frac{\pi}{2}$. | theorem Shakarchi_exercise_2_2 :
Tendsto (fun y => ∫ x in (0 : ℝ)..y, Real.sin x / x) atTop (𝓝 (Real.pi / 2)) := by
sorry | We have $\int_0^{\infty} \frac{\sin x}{x} d x=\frac{1}{2 * i} \int_0^{\infty} \frac{e^{i * x}-e^{-i * x}}{x} d x=\frac{1}{2 * i}\left(\int_0^{\infty} \frac{e^{i * x}-1}{x} d x-\int_0^{\infty} \frac{e^{-i * x}-1}{x} d x=\right.$ $\frac{1}{2 * i} \int_{-\infty}^{\infty} \frac{e^{i * x}-1}{x} d x$. Now integrate along the... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
364 | Shakarchi_exercise_2_13 | Shakarchi | import Mathlib
open Complex Filter Function Metric Finset
open scoped BigOperators Topology | Suppose $f$ is an analytic function defined everywhere in $\mathbb{C}$ and such that for each $z_0 \in \mathbb{C}$ at least one coefficient in the expansion $f(z) = \sum_{n=0}^\infty c_n(z - z_0)^n$ is equal to 0. Prove that $f$ is a polynomial. | theorem Shakarchi_exercise_2_13
{f : ℂ → ℂ}
(hf_diff : Differentiable ℂ f)
(hf_vanish : ∀ z₀ : ℂ, ∃ n : ℕ,
iteratedDeriv n f z₀ / (n.factorial : ℂ) = 0) :
∃ (p : Polynomial ℂ), f = fun z => Polynomial.aeval z p := by
sorry | Say that at least one of the coefficients of the Taylor series vanishes is the same as saying that for every $a \in \mathbb{C}$ there is $m \in \mathbb{N}$ such that $f^{(m)}(a)=0$.
Consider $A_n:=\left\{z \in \mathbb{C}: f^{(n)}(z)=0\right\}$ for each $n \in \mathbb{N}$. Note that:
$f$ is polynomial iff $A_n$ is not c... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
365 | Shakarchi_exercise_3_4 | Shakarchi | import Mathlib
open Complex Filter Function Metric Finset MeasureTheory Real Set
open scoped BigOperators Topology FourierTransform RealInnerProductSpace Complex Interval | Show that $ \int_{-\infty}^{\infty} \frac{x \sin x}{x^2 + a^2} dx = \pi e^{-a}$ for $a > 0$. | theorem Shakarchi_exercise_3_4 (a : ℝ) (ha : 0 < a) :
Tendsto (λ y => ∫ x in -y..y, x * Real.sin x / (x ^ 2 + a ^ 2))
atTop (𝓝 (Real.pi * (Real.exp (-a)))) := by
sorry | $$
x /\left(x^2+a^2\right)=x / 2 i a(1 /(x-i a)-1 /(x+i a))=1 / 2 i a(i a /(x-i a)+i a /(x+
$$
$i a))=(1 /(x-i a)+1 /(x+i a)) / 2$. So we care about $\sin (x)(1 /(x-i a)+$ $1 /(x+i a)) / 2$. Its residue at $x=i a$ is $\sin (i a) / 2=\left(e^{-a}-e^a\right) / 4 i$.? | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
366 | Shakarchi_exercise_3_14 | Shakarchi | import Mathlib
open Complex Filter Function Metric Finset MeasureTheory
open scoped BigOperators Topology Polynomial | Prove that all entire functions that are also injective take the form $f(z) = az + b$, $a, b \in \mathbb{C}$ and $a \neq 0$. | theorem Shakarchi_exercise_3_14 {f : ℂ → ℂ} (hf : Differentiable ℂ f)
(hf_inj : Function.Injective f) :
∃ (a b : ℂ), f = (λ z => a * z + b) ∧ a ≠ 0 := by
sorry | Look at $f(1 / z)$. If it has an essential singularity at 0 , then pick any $z_0 \neq 0$. Now we know that the range of $f$ is dense as $z \rightarrow 0$. We also know that the image of $f$ in some small ball around $z_0$ contains a ball around $f\left(z_0\right)$. But this means that the image of $f$ around this ball ... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
367 | Shakarchi_exercise_5_1 | Shakarchi | import Mathlib
open Complex Filter Function Metric Finset
open scoped BigOperators Topology | Prove that if $f$ is holomorphic in the unit disc, bounded and not identically zero, and $z_{1}, z_{2}, \ldots, z_{n}, \ldots$ are its zeros $\left(\left|z_{k}\right|<1\right)$, then $\sum_{n}\left(1-\left|z_{n}\right|\right)<\infty$. | theorem Shakarchi_exercise_5_1
(f : ℂ → ℂ) (hf : DifferentiableOn ℂ f (ball 0 1))
(hb : Bornology.IsBounded (f '' (ball (0 : ℂ) 1)))
(h0 : ¬ Set.EqOn f (0 : ℂ → ℂ) (ball (0 : ℂ) 1)) :
Summable (fun z : {z : ℂ // f z = 0 ∧ z ∈ ball (0 : ℂ) 1} =>
((analyticOrderAt f (z : ℂ)).toNat : ℝ) * (1 - ‖(z : ... | Fix $\mathrm{N}$ and let $D(0, R)$ contains the first $\mathrm{N}$ zeroes of f. Let $S_N=\sum_{k=1}^N\left(1-\left|z_k\right|\right)=$ $\sum_{k=1}^N \int_{\left|z_k\right|}^1 1 d r$. Let $\eta_k$ be the characteristic function of the interval $\left.\| z_k \mid, 1\right]$. We have $S_N=\sum_{k=1}^N \int_0^1 \eta(r) d r... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip |
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