# Tag Info

3

Examples aren't in mathematics. They are tools for learning mathematics, but are exterior to mathematical systems. There couldn't be a "mathematical" symbol for it for just that reason. Granted, you could use a symbol, but it would be just a symbol that represents a word, not an acceptable character in a well-formed formula.

3

As a first thought, I would say $f^{-1}$ is always injective, since a function is invertible if and only if it is injective and surjective. $f^{-1}$ is certainly invertible since its inverse is $f$, thus it is also injective. For the bounded question, consider the function f(x) = \left\{ \begin{aligned} &\dfrac{1}{1+(\frac{1}{x})^2} &&: x ... 2 A counterexample was given by David Mitra:f(x)=\begin{cases}x,\quad & x\in\mathbb{Q} \\ x+1 , \quad &x\notin\mathbb{Q}\end{cases}$$This is an everywhere discontinuous bijection, the inverse being$$f^{-1}(x)=\begin{cases}x,\quad & x\in\mathbb{Q} \\ x-1 , \quad &x\notin\mathbb{Q}\end{cases}$$0 Ramsey's theorem For any given number of colors c, and any given integers n1, . . . ,nc, there is a number, R(n1, ..., nc), such that if the edges of a complete graph of order R(n1, . . . , nc) are colored with c different colors, then for some i between 1 and c, it must contain a complete subgraph of order ni whose edges are all color i. The two ... 0 HINT: Let \theta any real number such that \frac{\theta}{\pi} is irrational. The metric space (e^{i n \theta})_{n\ge 0} is isometric to the subset (e^{i n \theta})_{n\ge 1} by the map z \mapsto e^{i\theta} \cdot z 1 I drew UG and added six more isolated vertices. Now, in G' K6 is a subgraph and therefore G' is nonplanar. 3 The point is that if p:X\rightarrow Y and q:Y\rightarrow Z are covering maps in this sense, but some z\in Z has infinitely many y_i\in Y with q(y_i)=z, then each y_i may have some neighborhood U_i\subseteq Y "evenly covered" by p (I take this to mean the inverse image of U_i is a union of parts each mapped isomorphically to U_i) but as ... 3 By a theorem of Darboux, a derivative has the Intermediate Value Property. So it cannot be positive at 0 and negative somewhere else without being 0 somewhere in between. 2 What about? \mathbb{R} \rightarrow \mathbb{R}, f(x)=e^{x^2} 4 I think you need some restrictions on A and B. For example, if k=2, suppose 27,243\in A and 81\in B. Then set y:=x_1^{27} and z:=-x_2^{27}, and obtain$$ |y-z|\le \frac{1}{27},\quad |y^3-z^3|\ge 1\quad\text{and}\quad |y^9-z^9|\le \frac{1}{243}. $$This is impossible for every pair y,z\in\Bbb{C}, since$$ ...

0

I would try to prove that the result is still false also in the case of a bounded double complex. Let's work in an abelian category $A$ with enough projectives, and let $F$ be a right-exact functor (for instance, you could think to the tensor functor over the category of modules). Let $0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0$ be a ses ...

6

Just a remark: Since both $A$ and $C$ are connected, if $A\cup C$ is disconnected, then $A$ and $C$ form a separation of their union. So there are closed sets $B$ and $D$ in $X$ such that $A\subseteq B\subseteq X\setminus C$ and $C\subseteq D\subseteq X\setminus A$. Since $C$ is closed in $X\setminus A$, there is a closed set $E$ such that $C=E\setminus A$. ...

3

This is not necessarily true. Consider the following bicomplex: $$K^{p,q} = \begin{cases} \mathbb{Z}, & q=-p \text{ or } q=-p+1; \\ 0, & \text{otherwise.} \end{cases}$$ The differentials are defined as follows: $d : K^{p,-p} \to K^{p,-p+1}$ (the vertical maps) is multiplication by $2$, while $\delta : K^{p,-p} \to K^{p+1,-p}$ (the horizontal maps) ...

2

Consider $\mathbb{R} \setminus \mathbb{R} = \varnothing$, which does not have the same homotopy type as $\mathbb{R} \setminus \{0\} \sim S^0$ even though $\mathbb{R}$ deformation retracts onto a point. For more complicated examples (because this one could look "pathological"), you can remove a half-line from $\mathbb{R}$ and get a contractible space (not ...

7

Consider $\sin(x^2)$. It is continuous and bounded but not uniformly continuous. This is because we can find arbitrarily small $\gamma$ so that $f(a)=-1$ and $f(a+\gamma)=1$. In other words if $\epsilon=1$ we cannot find a suitable $\delta$. This is because the function oscillates faster and faster as $|x|$ grows. On the other hand what is true is that a ...

2

Many problems can be reformulated to precisely ask for a fixed point of a function. A very simple example is the following. Suppose that $A$ is an $n\times n$ matrix, thought of as a function $A\colon \mathbb R^n\to \mathbb R^n$. Solving $Ax=b$ is an extremely important problem (I hope this does not require elaboration). Now, $Ax=b$ holds if, and only if, ...

2

For non-loops, neither implication holds. If we take the half-circle $\alpha \colon [0,1] \to S^1,\: \alpha(t) = e^{\pi i t}$, and look at $f(z) = z$, $$g = f\ast c_1 \colon z \mapsto \begin{cases} z^2 &, \operatorname{Im} z \geqslant 0 \\ 1 &, \operatorname{Im} z < 0,\end{cases}$$ and $h(z) = z^2$, then we have $f \simeq g \not\simeq h$, but ...

0

Claim: Given a set of $n$ points. Then these points lay on one line. Proof: Inductive Basis: Clearly, one point lays on one line. Inductive Hypothesis: Given a set of $k$ points. Then these points lay on one line. Inductive Step: Consider a set of $k+1$ points. Consider a subset of $k$ points. Then these lay on a line. Consider another subset of $k$ ...

0

$$f(x)=\begin{cases} \exp\left(\frac{1}{x^2-x}\right) &\mbox{ for } 0<x<1\\ 0 &\mbox{ for } x=1 \end{cases}.$$

2

As PhoemueX already said, you need a function, which has zero boundary values, but the derivative is non-zero at the boundary. In particular, you can take $$u(x) = \prod_{i=1}^d x_i \, (1-x_i).$$ Then, it is easy to see that $u(x) = 0$ for $x \in \partial[(0,1)^d]$, but $\nabla u(x) \ne 0$ for $x \in \partial[(0,1)^d]$. Since $u$ is smooth on $[0,1]^d$, ...

2

Restrict the usual covering projection from the real line to the circle(that is $x$ goes to $e^{2\pi ix}$) on the positive side of the line. $p:R_+\rightarrow S^1$ defined as $p(x)=(\cos2\pi x,\sin2\pi x)$ This is a local homeomorphism but not covering projection. As the point $(1,0)$ on the circle is not evenly covered by the projection map.

4

Hint: Consider $(X,Y) := (X,X)$ for a continuous random variable $X$. Show that $$D := \{(x,y) \in \mathbb{R}^2; x=y\}$$ has Lebesgue measure zero, but $\mathbb{P}((X,Y) \in D)=1$. Conclude that $(X,Y)$ is not absolutely continuous (with respect to the Lebesgue measure).

1

Take $X$ and $Y$ as being fully correlated bivariate normal random variables. Then the joint density is given by dirac masses as follows: $$f(x, y) = g(x)\delta_x(y)$$ where $g(x)$ is a standard normal density.

1

The answer to the first question is No. If this were possible for a given infinite set $X$, then every subset would be closed. So every subset would also be open, including one point subsets. That means that it' s the discrete topology.

3

If $X$ is Hausdorff and every subset of $X$ is compact, then every subset of $X$ is closed, and hence every subset of $X$ is open. So we necessarily get the discrete topology. Since $X$ with the discrete topology is itself compact, we must have that it's finite, thus contradicting your assumption. So no, as soon as $X$ is Hausdorff, and metric spaces are ...

2

Let $\tau$ be the usual topology of $\mathbb R.$ Define a finer topology $\tau_1$ as follows: $$\tau_1=\{X\subseteq\mathbb R:X\cap(0,\infty)\in\tau\}.$$ The function $f(x)=x+1$ is a continuous bijection, but not a homeomorphism, from $(\mathbb R,\tau_1)$ to itself.

1

Let $K=\{0\}\cup\left\{\frac1n:n\in\Bbb Z^+\right\}$ with the topology that it inherits from $\Bbb R$, let $\Bbb N$ have the discrete topology, and let $X=\Bbb N\times K$ with the product topology. Let $$A=\{\langle n,y\rangle\in X:n\text{ is odd}\}\;;$$ $A$ is countably infinite, so let $\varphi:\Bbb N\times\{1\}\to A$ be any bijection. Then $$f:X\to ... 2 The simplest - in the sense of least things to verify - examples that I can think of are linear operators on some infinite-dimensional vector spaces. For example, endow the space$$c_{00}(\mathbb{N}) := \bigl\{ x \colon \mathbb{N}\to \mathbb{C} : \bigl(\exists k\in\mathbb{N}\bigr)(n \geqslant k \implies x(n) = 0)\bigr\}$$with an \ell^p-norm for 1 ... 0 Consider the following Cayley Table of order 5:$$\begin{array}{c|ccccc} \ast & 1 & 2 & 3 & 4 & 5\\ \hline 1 & 1 & 2 & 3 & 4 & 5\\ 2 & 2 & 1 & 3 & 3 & 5\\ 3 & 3 & 3 & 1 & 4 & 5\\ 4 & 4 & 3 & 4 & 1 & 5\\ 5 & 5 & 5 & 5 & 5 & 1 ...

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