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## Hot answers tagged compactness

17

A compact set must be bounded. Otherwise we can take $\{ x_n \}_{n=1}^\infty$ such that $\| x_n \| \geq n$. This will have no convergent subsequence, which we can prove by showing that it has no Cauchy subsequence. A compact set must be closed. Otherwise we can pick a sequence which converges to a point in the closure which is not in the set. This will ...

7

They seem to be usually called KC-spaces (Kompact Closed), occasionally TB-spaces, and very rarely $J_1^\prime$-spaces. As you noticed, this class of spaces lies strictly between the T1-spaces and the Hausdorff spaces. I am unaware of any characterisation of them apart from the definition given. The closest thing of this kind I can think of is that a ...

5

I need a couple of preliminary results. Lemma $\mathbf 1$: Suppose that $X$ is $T_1$ and weak Hausdorff. Let $K$ be a compact Hausdorff spaces and $f:K\to X$ continuous; then $f[K]$ is not just compact, but also Hausdorff. Proof: Let $x$ and $y$ be distinct points of $f[K]$, and let $H_x=f^{-1}[\{x\}]$ and $H_y=f^{-1}[\{y\}]$; $H_x$ and $H_y$ are ...

4

Every compact metric space is separable, so let $D=\{x_k:k\in\Bbb N\}$ be a dense subset of $X_1$. For each $n\in\Bbb N$ let $D_n=\{x_k\in D:k\le n\}$; $D_n$ is finite, so there is an isometry $f_n:D_n\to X_2$. For each $k,n\in\Bbb N$ with $k\le n$ let $y_k^n=f_n(x_k)$, so that $f_n[D_n]=\{y_k^n:k\le n\}$. Note that for any $k,\ell,n\in\Bbb N$ with ...

4

The following result is available: Let $(X,\lVert\cdot\rVert)$ be a Banach space and $S$ a subset of $X$. Then $S$ is compact if and only if the following two conditions hold: $S$ is bounded; for each positive $\varepsilon$, there exists a finite dimensional space $F=F(\varepsilon)$ such that for all $x\in S$, we have $d(x,F)=\inf\{\lVert ... 3 Since$\operatorname{cl}_{\beta\omega}Y\subseteq\operatorname{cl}_{\beta\omega}X$, you might as well work in the subspace$\operatorname{cl}_{\beta\omega}X$. In that case you can assume without loss of generality that$X=\omega$. Fix$p\in(\operatorname{cl}_{\beta Y}Y)\setminus Y$; if we view$p$as an ultrafilter on$\omega$, then$Y\in p$, and there is an ... 3 Not quite. A metric space space can be complete without being compact (e.g.,$\mathbb R$with the Euclidean topology). For a metric space, completeness + total boundedness = compactness. 3 Hint: Fix any element of an open cover of$K$that contains$p_0$. Finitely many remain out there. 3 If$(X,\tau_X)$and$(Y,\tau_Y)$are homeomorphic, then there is a bijection between the two spaces. Namely, there is a function$f\colon X\to Y$which is a bijection, and satisfies that$f[U]$is open if and only if$U$is open (for$U\subseteq X$). Now ask yourself, is there a bijection$f\colon\Bbb R\to\Bbb Z$? 3 If$X$is metric and not compact, then$X$has an infinite closed discrete subset$D$. By passing to a subset if necessary, we may assume that$D=\{x_n:n\in\Bbb N\}$is countably infinite. Now define$f:D\to\Bbb R:x_n\mapsto n$, and apply the Tietze extension theorem to get an unbounded continuous real-valued function on$X$. 3 In order to prove that the space$\omega_1$is locally compact, you have to prove that every point in$\omega_1$has a compact neighborhood. If$\alpha\in\omega_1$then the one-point set$\{\alpha\}$is obviously compact, but it is not necessarily a neighborhood of$\alpha$; namely, it's not a neighborhood if$\alpha$is a limit ordinal, such as$\omega$. ... 2 You can write each$z_n$in that way. Since$A$is compact, there exists a convergent subsequece$(x_{n_j})$. Now, the subsequence$(y_{n_j})$, also admits a convergent subsequence$y_{n_{j_k}}$. So the sequence$(z_n)$, admits a convergente subsequence$z_{n_{j_k}}$, and$AB$is compact. 2 I think you are on a right way, and in your proof you can use the fact that there is no infinite decreasing sequence of ordinals. 2 You can explicitly write down a parametrization of$K$, showing that it's a continuous image of a compact set. But here is a general argument. Claim If$A\subset \mathbb R^n$is compact and$C$is the smallest convex set containing$A$, then$C$is compact. Proof. The set$C$, also known as the convex hull of$C$consists of all finite convex ... 2 Homeomorphic spaces share their topological properties. A discrete space with more than one element is not connected; the real line with the usual topology is connected. The compact subsets of a discrete space are finite, the real line has plenty of infinite compact subsets. Sequences with pairwise distinct terms are not convergent in a discrete space, ... 2 I think the only limit point of the first set$S_1$but not in$S_1$is$(0,0)$. Since for any$p=(u,v)\neq (0,0)$and$p$is not an element of the first set$S_1$, let$r^2=u^2+v^2$, we can find sufficient large$x$, s.t$e^{-x}<r$. Then we can find a neighborhood of$p$, which doesn't intersect$S_1$. Hence$p$is not a limit point. Since$(0,0)$is ... 2 An open cover of$(-2,2)$can only be broken down into$(-2,0) \cup (0,2)$if those two open sets are included in the cover, not to mention that$(-2,0) \cup (0,2)$fails to cover$(-2,2)$to begin with (what contains$0$?). You are right when you say that open sets of$\mathbb R$are not compact (well the empty set is). To show this, you would have to ... 2 For instance,$f:[-1,1] \rightarrow \mathbb R$defined as$f(x)=1/x$if$x\neq 0$and$f(0)=0$is defined on a compact domain$[-1,1]$but it is not bounded. Recall the Weierstrass theorem: "Every continuous function on a compact domain has at least one maximum and one minimum" So negating the above statement we obtain that: "No maximum or minimum and ... 1 consider for example $$f:[0,1]\to \Bbb R\\ f(x) = \begin{cases} \frac 1x &\text{if} & \frac 1x\in\Bbb N, x\neq 0 \\ 0 &\text{otherwise} \end{cases}$$ which is unbounded on$[0,1]$. 1 Generalization of completeness to non-metric spaces goes through the concept of uniform spaces. 1 No. Here is a counterexample, inspired by the answer to this question: If$V \subset H$compact, is$L^2(0,T;V) \subset L^2(0,T;H)$compact too? Take$v\in H^1$,$T=\pi$, $$\phi_n(t) = \sin(n t)v.$$ Since$\sin(n \circ)$converges weakly but not strongly to zero in$L^2(0,T)$, it follows that$\phi_n$converges weakly to zero in$L^2(0,T;H^1)$. If ... 1 HINT: For any$p\in\Bbb R^2$and any$\epsilon>0$, the set$\Bbb R^2\setminus\operatorname{cl}B(p,\epsilon)$is open. (Here$B(p,\epsilon)$is the open ball of radius$\epsilon$and centre$p$.) Added: This doesn’t arise in your specific example, but in general you need more than that$p$is in the boundary of the compact set: you need it to be a limit ... 1 Let$X$be a compact Hausdorff space and$p\in X$apoint such that$Y:=X-\{p\}$is not closed. For each$x\in Y$there exist disjoint open sets$x\in U_x,p\in V_x$. Then the$U_x$cover$Y$. Assume there is a finite subcover$U_{x_1}\cup\ldots\cup U_{x_n}$. Then this subcover misses the open set$V_{x_1}\cap \ldots\cap V_{x_n}$which contains$p$ans must be ... 1 Let$p=(0,-1) \in \mathbb{R}^2$, and consider the collection of open balls$\{B(p,2-\frac{1}{n})\}_{n \in \mathbb{N}}$1 Let's do it according to the most standard definition of compactness: A subset$K$of a metric space$X$is compact if every open cover has a finite subcover. Let$\{U_i\}_{i\in I}$be an open cover of$K$. The$p_0\in U_{i_0}$, for some$i_0\in I$. But as$U_{i_0}$is open, then a whole ball$B(p_0,\varepsilon)\subset U_{i_0}$. Next, as$p_n\to p_0$, ... 1$K_1$might have lots of points in common with other$K_\alpha$s, indeed it must since every finite sub-collection (such as every intersection of two sets) has nonempty intersection. The point is that we assume that no point of$K_1$is in every$K_\alpha$(by way of contradiction). I suppose you could assume (again by way of contradiction) that there was ... 1 For each$a\in\mathbb R$, let$K_a=[a,\infty)$. Note that$K_a$is a family of closed subsets of$\mathbb R$in which every finite subcollection has non-empty intersection, however$\bigcap K_a=\varnothing$. So you need compactness at some point. 1 I really don't like the way the proof is presented in Rudin, and I'd go as follows: suppose$\;\bigcap_\alpha K_\alpha=\emptyset\;$, but then fixing$\;\alpha_0\;$we get (de Morgan): $$K_{\alpha_0}=K_{\alpha_0}\setminus\emptyset=K_{\alpha_0}\setminus\bigcap_\alpha K_\alpha=\bigcup_\alpha\left(K_{\alpha_0}\setminus K_\alpha\right)$$ But ... 1 If the conclusion fails, then there is, in$X$, a first element$\beta$such that$]\leftarrow,\beta]$is not compact. Let$C$be an open cover of$]\leftarrow,\beta]$with no finite subcover, and let$U$be an element of$C$that contains$\beta$. Of course,$U$doesn't include all of$]\leftarrow,\beta]$, so, by definition of the topology, it includes ... 1 Prove by induction on$\beta$that$L(\beta) = \{x: x \le \beta \}$is compact, for all ordinals$\beta$. This is clear for$\beta = 0$, where$L(0) = \{0\}$and if$\beta+1$is a successor, then$L(\beta+1) = L(\beta) \cup \{\beta+1\}$, so if$L(\beta)$is compact, so is$L(\beta+1)$. So assume$L(\alpha)$is compact for all$\alpha < \beta\$ and ...

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