# Tag Info

## Hot answers tagged compactness

9

No, since the image of a compact map by a continuous map is compact and the projection $p:X^n\rightarrow X$ is continuous and surjective.

9

The claim is true. A proof is reported by Theorem 1 in this paper by A. H. Stone (in case you’re wondering, this is not the Stone–Weierstrass-theorem guy), which I outline below. Let $B$ denote the boundary of $f^{-1}(\{y\})$ and suppose, for the sake of contradiction, that it is not compact. Then, there exists a sequence $(x_n)_{n\in\mathbb N}$ in $B$ such ...

8

open cover (for a given topology $\tau$): a collection of open sets $U_{\alpha}$ (open with respect to the topology $\tau$) such that $$X \subset \underset{\alpha \in A}{\bigcup} U_{\alpha}$$ open set: A set in the topology $\tau$ of the space $X$. compact (for a given topology $\tau$): every open cover of $X$ (with respect to the topology $\tau$) has a ...

6

Consider the subset of $\mathbb R$ $$\{ \frac{1}{n} | n\in \mathbb N^*\} \cup \{0\}\;$$ which inherits a metric structure from the standard metric of $\mathbb R$. The sets $\{1/n\}$ are countably infinitely many clopen subsets.

6

$X=(0,1)$ with Euclidean metric. Then $X$ itself is closed and bounded yet not compact.

5

Your confusion here is that there are two different topologies in play. We have the topology on $X$ and the subspace topology on $A$. The lemma says that $A$ is compact with respect to its subspace topology if and only if "$A$ is compact with respect to the whole topology". I put this last part in quotes because this second idea of "compactness" isn't ...

5

You should use elements of the topology (open sets) to cover your space.

5

Also in your lecture notes should be: Let $X$ be a set, $(X,\mathcal{T})$ a topological space. Then a collection $\mathcal{A} \subset \mathcal{P}(X)$ is called an open cover if: (i) $\mathcal{A}$ is a cover, i.e. $\forall x \in X, \exists A \in \mathcal{A}$ such that $x \in A$. (ii) Elements of $\mathcal{A}$ are open, i.e. $\mathcal{A} \subset ... 5 If the$n$real-valued continuous functions$f_1, \ldots, f_n$separate points of$K$, then$(f_1, \ldots, f_n)$is a homeomorphism from$K$to a compact subset of$\mathbb R^n$. But not every compact metric space is homeomorphic to a compact subset of$\mathbb R^n$. For example, let$K$be the Hilbert cube. For each$k$,$K$has a subset$S_k$... 4 The collection$\mathcal{A}$may be infinite, in which case it need not have a smallest element. For instance, taking$X=[-1,1]$, you could have$\mathcal{A}=\{[-r,r]:0<r\leq 1\}$. This collection is simply ordered, but it has no smallest element. 3 Hint:$S_k=[-k,k]$[edit after the question was changed] If you want$S_k\subset\tilde{S}$for some compact$\tilde{S}$then let$S_k$be the unit line segment intersecting the origin of the plane at angle$\theta=2\pi/k$. 3 You cannot really construct it, as such. You can define it, and prove its existence (using the Axiom of Choice) but you cannot give a concrete, definable example of a point in the remainder$\beta\mathbb{R}\setminus\mathbb{R}$. It is common to define the half-line$\mathbb{H} = [0,\infty)$and consider$\beta\mathbb{H} \setminus \mathbb{H}$, because one can ... 3 Since$X$is compact, there exists some finite$J\subseteq I$such that$\bigcup U_j=X$. That is the mistake: The open cover that covers$A$was not assumed to cover all of$X$, so it's not an open cover of$X$. If it doesn't cover$X$then it can't have a finite subset that covers$X$. 3 If the logic you're using admits excluded middle, you could convert the hypothesis to $$⋀ Φ ∧ ⋀ ¬ ψ ⊢ ⊥.$$ Then Compactness would imply some finite subset$\phi₁ ∧ … ∧ φ_n ∧ ¬ ψ₁ ∧ … ∧ ¬ ψ_m$of the left hand side entails a contradiction, and from there you can move$\psi₁$through$\psi_m$back to the right hand side. 3 If you are familiar with degrees, one can argue as follows. The identity selfmap$z\mapsto z$of the circle has degree$1$. The squaring map$z\mapsto z^2$has degree$2$. The function$(g(z))^2$is the composition of$g$and$z\mapsto z^2$. Under composition the degree multiplies, i.e., the degree of$g$would have to be$\frac{1}{2}$in order for the ... 3 Let$n \in \mathbb{N}$, and define$ f_n(x) = \left\{\def\arraystretch{1.2}% \begin{array}{@{}c@{\quad}l@{}} 0 & \text{if } 0 \leq x \leq 1/2\\ 2nx - n & \text{if } 1/2 < x \leq (n + 1)/(2n)\\ 1 & \text{if } (n + 1)/(2n) < x \leq 1\\ \end{array}\right.$Then$\{f_n\}$is a Cauchy sequence whose limit is$ f(x) = ...

3

No. Say $K$ is the "Hilbert cube", that is, the set of all sequences $x=(x_1,\dots)$ with $$0\le x_n\le 1/n$$for all $n$, and the metric $$d(x,y)=\left(\sum(x_n-y_n)^2\right)^{1/2}.$$The question is equivalent to asking whether there exists $N$ and a continuous injective mapping from $K$ to $\Bbb R^N$. And the answer to that is no. Let $$K_N=\{x\in ... 2 Some remarks on this, without striving for comprehensiveness: Compactness and sequential compactness do not imply each other in general topological spaces. For example, the set of countable ordinals with the order topology is not compact, but it is sequentially compact, whereas the space \{0,1\}^{[0,1)} with the product topology (each factor space being ... 2 It’s true in any metric space. If X is metric, then E is compact if and only if E is countably compact. If E is not countably compact, then E has an infinite discrete subset D that is closed in E. Let D=\{x_n:n\in\Bbb N\}, and define f:D\to\Bbb R:x_n\mapsto n; f is continuous on D,so by the Tietze extension theorem there is a g\in ... 2 A space X is called pseudocompact when every real-valued continuous function on X is bounded. This is a well-studied notion. Indeed a compact space, or even a countably compact space (every countable open cover has a finite subcover) is pseudocompact. A classical theorem: for normal and T_1 spaces, countably compact and compact are equivalent. The ... 2 In any metric space a compact sub space is closed and bounded. (The reciprocal is not true in general). Take 0\neq x\in F and n\in\Bbb{N}. One has$$\|n\cdot x\|=n\cdot\|x\| And the sequence $y_n=n\cdot x\in F$ is not bounded ; therefore $F$ is not bounded henceforth not compact.

2

Such functions are called proper. In general, they have the property that preimages of compact sets are compact. For the first question, note that a space $A$ is Hausdorff if and only if for any $a_1,a_2$ we have closed sets $F_1,F_2$ such that $a_i\notin F_i$ and $F_1\cup F_2=A$, and that in a Hausdorff space, you can separate any two disjoint compact ...

2

For $1$, start with any bounded sequence $\{ f_n \}$ in $C[0,1]$ that has no convergent subsequence. Then $F_n\int_{0}^{x}f_n(t)dt$ gives you a bounded sequence $\{ F_n \} \subset C^1[0,1]$ whose image under $T$ is $\{ f_n \}$. For $2$, the same technique holds. Start with a bounded sequence in $C^1[0,1]$ with no convergent subsequence, and integrate. For ...

2

The conclusion I want to jump to is that the infinite union is compact. But, that is not true. Suppose $S_n = \{1,\frac1n\}$ and $\bar S = \{\frac1n: \forall n \in \mathbb N\} \cup \{0\}$ $\bar S$ is compact. $\bigcap S_k = \{1\}$ i.e. non-empty. There exists a sequence $\in\bigcup S_k$ that converges to $0.$ $0 \notin \bigcup S_k$ $\bigcup S_k$ is ...

2

Embed $K$ into a product $[0,1]^I$ for some $I$ ($I$ can have the size as a base for $K$), say via $i: K \rightarrow [0,1]^I$, where $i: K \rightarrow i[K]$ is a homeomorphism. Then consider the maps $p_i:= \pi_i \circ i \circ f$ for all $i$. The $p_i$ are among the functions in $S$. Then $g_i = \pi_{p_i}: \overline{e[X]} \rightarrow [0,1]$ restricted to ...

2

An open cover is a collection of open sets whose union is $X$. Thus it is a subset of $\tau$ to begin with. Since every subset of $\tau$ is finite, then so is every open cover.

2

HINT: For each $\alpha\in A$ let $U_\alpha=X\setminus C_\alpha$; then $\{U_\alpha:\alpha\in A\}$ is an open cover of $X\setminus U$. Since $X\setminus U$ is closed, it is also compact. Can you finish it from here?

2

Consider $X = \mathbb{Z}$ with a metric $d$ such that $d(m, n) = 1$, if $m \neq n$ and $d(n, n) = 0$. $d$ is a metric. Take any infinite subset $A$ (for example, $\mathbb{N}$). Then $A$ is open (and closed) and it is also bounded (because the maximum distance is 1). Moreover, $A$ is not compact, because the family of sets defined by $\{\{x\} \mid x \in ... 2 Since the$f$is continuous and bijective, it is a homeomorphism if and only if it is closed. But closed subset of a compact space are compact, image of compact is compact, and compact subsets of a Hausdorff space are closed. 2 Seems to me the definition should require just that$h|_X$be a homeomorphism of$X$instead of requiring it to be the identity on$X$. Examples exist with that revised definition, but I can't think of one. With the given definition it's easy to give an example. For complex numbers$a$and$b$let$(a,b)$denote the open line segment with endpoints$a\$ and ...

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