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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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