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8

Hint: Consider the (continuous!) function $g:X \to \Bbb R$ given by $$ g(x) = d(x,f(x)) $$ Why must $g$ achieve its minimum? To do this in a manner similar to the way you originally planned: for each $n \in \Bbb N$, define $U_n = \{x \in X: d(x,f(x)) > 1/n\}$. Take a finite subcover.


6

This answer is preliminary. Now I am looking in the papers and soon I shall extend the answer. I am related with the subjects, but I am not a specialist. I looked into papers and can say you the following. If you are so interested and want to have a perfect answer, you may ask Hans-Peter Künzi (use this e-mail: Hans-peter.Kunzi (at) uct.ac.za). So, IMHO, ...


6

HINT: If $S$ is a theory, and $\psi$ is a statement such that $S\not\models\psi$, then $S\cup\{\lnot\psi\}$ is consistent.


4

Hint Assume by contradiction that this is not true. Then for each $n$ you can find $x_n$ so that $d (x_n, f (x_n)) \le \frac{1}{n}$. Now $x_n$ has a cluster point $y$ (Why?). What is $f(y)$?


3

$T \models \phi$ means that every model of $T$ is also a model of $\phi$. The compactness theorem states that if every finite subset of $T$ has a model, then $T$ has a model. $T$ is consistent means that there is some formula $\phi$ such that $T \not\vdash \phi$ where $T \vdash \phi$ means that $\phi$ has a proof using axioms drawn from $T$. The soundness ...


3

In this answer the following lemma is proved: Lemma Let $X$ be a Hausdorff space and $C \subset X$ have a compact neighbourhood $K$. Then $C$ is a component of $X$ if and only if $C$ is a component of $K$ Also: Let $X$ be a compact Hausdorff space, $Y$ an open subspace and $Z$ a closed subspace. Let $C$ be a connected subset of $Y \cap ...


3

Ok, if $P$ is a polynomial, note $\|P\|$ the max of the absolute values of its coefficients and $V(P)$ the set of the roots of $P$. Then you have the following results : if $P$ is of degree $n$, for each $\varepsilon > 0$ there exist an $\eta >0$ such that for each $Q$ of degree $n$, $\|P-Q\| < \eta$ implies $V(Q)\subseteq V(P)+B(0,\varepsilon)$, ...


3

$X=[0,1]$ with the usual metric, $\phi(x)=\sqrt x$.


3

Revised and corrected to match the edited question. The statement is false. Let $X=[0,2]$ with the usual topology, and let $x=0$. Then $$X\setminus\{x\}=(0,2]=(0,1)\cup[1,2]\;,$$ where $(0,1)$ and $[1,2]$ are disjoint and connected, and $[1,2]$ is compact.


3

This is false. For example, $\mathfrak{sl}_2(\mathbb{R})$ is a semisimple Lie algebra, but $SL_2(\mathbb{R})$ isn't compact. This is also false, and a counterexample to 1 also provides a counterexample here. For example, $U(1) \times SL_2(\mathbb{R})$ isn't compact. This is also false, and a counterexample to 2 also provides a counterexample here. The ...


3

As noted in the comments, this answer shows that $f$ is closed if $R$ is closed in $X\times X$, i.e., that (3) implies (2). Now suppose that $f$ is closed. Note first that since $X$ is Hausdorff, $\{x\}$ is closed for each $x\in X$, and therefore $\{f(x)\}$ is closed for each $x\in X$. And $f$ is a surjection, so $\{y\}$ is closed for each $y\in Y$. Now ...


2

Use Holder's inequality: $$\int_x^y |u(t)| \, dt \le \left( \int_x^y 1 \, dt \right)^{1/p'} \left( \int_x^y |u(t)|^p \, dt \right)^{1/p} \le |x-y|^{1/p'} \|u\|_p.$$


2

For the case of finite measure space see chapter IV section 8 theorem 18 in Linear Operators. General theory. Volume 1 , N.Dunford, J. T. Schwartz For general case you should recall that $L_\infty$ is a $C(K)$-space for some weird Hausdorff compact space $K$ and apply Arzela-Ascoli theorem.


2

First of all, $S$ is bounded. Indeed, equip $(M_n(\Bbb R)\subset )M_n(\Bbb C)$ with a norm $\|\cdot\|$ subordinate to some norm $|\cdot|$ on $\Bbb C^n$. If $\lambda\in\Bbb C$ is an eigenvalue of $M\in M_n(\Bbb C)$ then $|\lambda|\leq\|M\|$, and thus, since $X$ is bounded, so is $S$. $S$ is also closed. Indeed, if you take a sequence $s_k$ of elements of $S$ ...


2

First, $\mathcal F$ is needlessly complicated. Let $f_x = f(x,\cdot)$. Since $|x| \le 1$, $f_x(y) \ge 0$ for all $y$. If $y_n \to \infty$ then $y_n \neq x$ for almost all $n$, and since $|x|<1$ and $y_n$ is bounded, $f_x(y_n)$ can't converge to $0$ anyway. Then you can see that $f_x(y_n)$ converges to $L$ if and only if $L > 0$ and $|x-y_n|$ converges ...


2

My idea how to start is so good that a start becomes a finish. :-) Put $$\alpha=\frac {\inf\operatorname{supp} f}2,\beta=1-\frac{1-\sup\operatorname{supp} f}2.$$


2

It seems the following. At first we consider compactness. (i) is called compactness, (ii) is called countably compactness, (iii) is called sequentially compactness. It is easy to check that a space is countably compact provided it is sequentially compact or compact. In particular, each compact space which is not sequentially compact (for instance, ...


2

What you have proved is essentially that $f(K)$ is compact. Then, as any compact in the real line, it is closed and bounded. In particular, it has a maximum $M$ and a minimum $m$, so $f(K)\subset[m,M]$. However, $m\in f(K)$, so there's $a\in K$ with $m=f(a)$; similarly, there is $b\in K$ such that $f(b)=M$. Therefore $f(K)\subset[f(a),f(b)]$. As far as ...


2

From the fact that $S$ is not closed, you should have concluded that $S$ is not compact. Connectedness (in fact, path-connectedness) can be verified without explicitly rewriting $S$ in a more comprehensible way (i.e., without recognizing that $S=[0,1)$. Note that $S$ is of the form $S=\{\,f(x)\mid x\in\mathbb R\,\}$ where $f$ is a continuous function. Then ...


2

I don't have the book handy, so I'll write down the proof in my own notation. The proof is as follows: suppose $K$ is compact relative to $X$. We want to show it is compact relative to $Y$: Take any open cover of $K$, $\{U_\alpha \mid \alpha \in A\}$. where all $U_\alpha$ are open in $Y$. As $Y$ has the subspace topology with respect to $X$, for every ...


2

Let $\mathcal A_1$ be the collection of all sets $E \subseteq X$ such that there exists a compact $G_\delta$ set $K$ with $x \in K \subseteq E$. Let $\mathcal A_2$ consist of the complements $E^c = X \setminus E$ for $E \in \mathcal A_1$. Let $\mathcal A = \mathcal A_1 \cup \mathcal A_2$. claim $\mathcal A$ is a sigma-algebra. Proof: $X \in \mathcal A$ ...


2

Problem: Can we write one to one, onto and continious from the circle $S_1$ to $\mathbb R$ ? The answer is no and it can be shown that if we extract one point from $S_1$, it is possible. Now the idea is that instead of extractinting one point from $S_1$, let's add one point to $\mathbb R$ so that we can write the map from $S_1$ to $\mathbb R \cup {a} $. ...


1

How $f(x,y)$ is continous in that set (because $x,y>0$). The set of the $(x,y)$ with $f(x,y)\leq \gamma$ is the preimage of the set $(-\infty,\gamma]$ (which is closed) and therefore is closed.


1

Let $\kappa$ be any uncountable cardinal. Let $D_\kappa$ be the discrete space of power $\kappa$, let $p$ and $q$ be distinct points not in $D_\kappa$, and let $X=D_\kappa\cup\{p,q\}$. Let $P_0$ be a countably infinite subset of $D_\kappa$, and let $Q_0=D_\kappa\setminus P_0$. Finally, let $\{P_1,Q_1\}$ be a partition of $D_\kappa$ into two sets of power ...


1

A continuous real-valued function on a compact set assumes a maximum and a minimum. So at some $x=a$, $f$ assumes the minimum of $f$ over $K$, and at some $x=b$, $f$ assumes it maximum over $K$. Then $f[K] \subseteq [f(a), f(b)]$ is obvious. We have garantueed equality when $K$ is connected, e.g. But this is not a necessary condition; this would be hard to ...


1

Suppose $U$ is an open subset of a locally compact Hausdorff space $X$, $K\subset U$ and $K$ is compact. Then there is an open set $V$ with compact closure such that $$K\subset V\subset \bar{V}\subset U.$$ This is Theorem 2.7 in Rudin's Real and Complex analysis. To prove it first you need to show (the easy) fact that in a Hausdorff space compact subsets ...


1

It seems the following. If $|b|\le 1$ then in order to converge, the series for $x$ should have only finitely many non-zero coefficients $d_k$. Moreover, for each $b$ the set $\Bbb N_b=\{k\in\Bbb N:\operatorname{arg} b^{-k}\in [-\pi/3;\pi/3]\}$ is infinite. Put $x_n=\sum_{k\in\Bbb N_b\cap [1,n]} \frac{1}{b^k}$. Since $\operatorname{Re} b^{-k}\ge ...


1

$B$ is always compact. Let $\mathscr{U}$ be an open cover of $B$. $A_0\subseteq B$, and $A_0$ is compact, so some finite $\mathscr{U}_0\subseteq\mathscr{U}$ covers $A_0$. Let $V=\bigcup\mathscr{U}_0$; $V$ is an open nbhd of the compact set $A_0$, so there is an $n\in\Bbb Z^+$ such that $A_n\subseteq V$. Let $K=\bigcup_{k=1}^nB_k$; then $K$ is a compact ...


1

Let $X=[0,1)$ with the given metric, and let $Y=[0,1]$ with the usual metric. If $(x_n)$ is a sequence in X, then it has a subsequence $x_{n_k}$ which converges to $a \in Y$ since Y is compact. 1) If $a\ne1$, then $x_{n_k}$ converges to $a$ in X since $\;\;\;|x_{n_k}-a|<\epsilon\implies d(x_{n_k},a)=\min\{|x_{n_k}-a|, 1-|x_{n_k}-a|\}<\epsilon$. 2) ...


1

Ok for $M$ bounded. But an argument like: $$\forall z\in M, z\in \overline B(0,1)=\{z\in\mathbb Z\mid |z|\leq1\}$$ would have been better. But anyway, you proved that $M\subset [-1,1]\times [-1,1]$ which is good too. But your set is not compact because $0\notin M$. Indeed, $0\in \overline M$ but not in $M$.



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