# Tag Info

7

Assume $X$ is Hausdorff. If every subspace is compact then every subspace is closed. So the topology is discrete. Now take the cover by singletons. If this has a finite subcover then the space must be finite.

7

For sure. Consider $X = \{a, b\}$ with topology $\tau = \{\emptyset, \{a\}, X\}$. Note that $(X, \tau)$ is not Hausdorff and that $\{a\}$ is compact (the only open covers for $\{a\}$ are already finite), but not closed.

6

Take $f(x) = \sin x$. Then $f: \mathbb R \to [-1,1]$. There are many other similar functions - for example $\frac 2 \pi \tan^{-1} x$ maps $\mathbb R \to (-1,1)$. The point is that whilst it is true that continuous functions take compact sets to compact sets, it is not generally true that the pre-image of a compact set is compact

6

Take $X$ to be the unit circle (not disk) and $f$ a non-trivial rotation. For an example in the real line, take $X=[-2,-1] \cup [1,2]$ and $f(x)=-x$. What fails in both cases is that $X$ is not convex.

5

Let $X=\{-1,1\}$ and let $f(x)=-x$.

5

If $X$ is a compact metric space, then $X$ itself is a compact subset of $X$, so the property $A$ must hold for $X$. If $X$ is not compact, then let $A$ be the property that a subset of $X$ is compact. Then $A$ holds for any compact subset of $X$, but $A$ does not hold for $X$ itself.

5

I agree with J. Loreaux's commrent above (“Your argument doesn't really work”), and I'd go farther: to me, your argument makes no sense whatever, for several reasons: You say “If the every compact set on a metric space is not bounded, then …”. This is at least confusingly stated. You want to prove that every compact set is bounded. You seem to be trying ...

4

Continuity preserves compactness. That is if $f \colon X \to Y$ is a continuous function betweeen topological spaces and $K \subset X$ is compact then the image $f(K)$ is compact. Addition $(-+ -) \colon \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ is a continuous function and if $A$ and $B$ are compact then $A \times B$ is compact in $\mathbb{R} \times ... 4 It is better to write ; for$n\in \mathbb{N}$, ($y_n\in f(K)$) there exists$x_n\in K$such that$y_n=f(x_n)$, to avoid$f^{-1}$... 4 An alternative approach to your problem: First, show that the function$f:(\Bbb R,\rho) \to (\Bbb R, |\cdot|)$taking$x \mapsto x$is continuous. Now, the continuous image of a compact set is compact. So, suppose that$\Bbb R$were compact under$\rho$. Then since$f$is continuous,$f(\Bbb R) = \Bbb R$would be compact under$|\cdot|$. Since this is ... 4 To generalize Mathmo123's example: since$f$is bijective we may as well consider$Y$to be the same set as$X$, but with a weaker topology (i.e. some of the open sets of$X$are no longer open). Then$f$is still continuous, but no longer a homeomorphism. So what the theorem is saying is that you can't weaken a compact topology and have it be Hausdorff, ... 3 No. It follows from the fact that a function can have only countably many strict local maxima and minima. For proof, see Countability of local maxima on continuous real-valued functions Part 1. Although continuousness is mentioned in the question, it isn't needed. 3 Every metric space$(M,d)$determines the associated topological space$(M,\mathcal{O})$whose open sets$U\in\mathcal{O}$are the unions of open balls of the metric space. Compactness of a metric space is a topological property, meaning that the metric space is compact iff the associated topological space is compact. Two homeomorphic topological spaces have ... 3 As said in comments, note that the structure of$\mathbb{Z}$like order or ring structure is completely irrelevant from topological view.$\mathbb{Z}$is just the countable discrete space, same as say$\mathbb{N}$or$ω$. Its one-point compactification is just$(ω + 1)$, i.e. the convergent sequence (as a space). And it is easy to find a convergent sequence ... 3 Each finite set$X$is compact. Using the discrete topology on$X$each map is continuous. So each permutation without a fixpoint will do the job. For$|X|>1$we always have permutations without a fixpoint. By the way: The discrete topology is just the induced topology if you consider finite subsets of$\mathbb{R}^n$. 3 Edited: I see why this is false but in general, why every closed subset of a compact set is compact? Another proof: Let$S \subset T$be a closed set, where$T$is compact. Let$\{\mathcal{U}_\alpha\}$be an open cover of$S$. Then$\{\mathcal{U}_\alpha\} \cup \{S^c\}$, where$S^c$is the complement of$S$w.r.t. to$X$, covers$T$. Since$T$is ... 3 Take the unit circle in$\mathbb{R}^2$and the map f, as f(x) going to its diametrically opposite point. This map is continuous but has no fixed point. 3 ADDED: There's actually a counterexample way simpler than the one below: $$[0,1]=\left[\frac12,1\right]\cup\bigcup_{n\in\mathbb N}\left[0,\frac12-\frac{1}{4n}\right).$$ ORIGINAL ANSWER: Let$X=$the Alexandroff one-point compactification of$\mathbb R$, with the extra point denoted as$\omega$. Let$S_0=\{\omega\}\cup(-\infty,0]\cup[1,\infty)$and ... 3 Fact: the uniform limit of a sequence of continuous functions is continuous. The problem is that the constant$M$depends on$n$. If$f_{n_k}$is a uniformly convergent subsequence, then for any$a\in [0,1)$, we have$f_{n_k}(a)\to 0$while$f_{n_k}(1)=\sin(1)$. We conclude that the potential limit function is not continuous, hence from the "fact" we reach ... 3 You''ve generated all examples. A corollary of the Peter-Weyl theorem is that every compact (Hausdorff) group is a closed subgroup of a product of$U(n)$s. 2 How can we find such an example? Since$f$is a bijection, we may assume wlog that$X$and$Y$are the same set, just with different topologies (and then$f$is just the identity map). The fact that$f$is continuous then means that every$Y$-open set is also an$X$-open set. To prevent$f$from being a homeomorphism, the converse should not hold, i.e. not ... 2 Yes. Take a compact Hausdorff space with topology$\tau$and weaken the topology, i.e. take any topology$\tau'$strictly weaker than$\tau$, so that there is some set$U$that is open in topology$\tau$but not in$\tau'$. But$U^c$is still compact in$\tau'$(because any open cover for$\tau'$is still an open cover for$\tau$). So$U^c$is a compact ... 2 I assume you are using the fact that$X$is compact if every sequence in$X$has a convergent subsequence with a limit in$X$. If that is the case, then what you wrote is not completely correct. You cannot just take two sequences in$A$and$B$and show that their sum has a convergent subsequence. What you need to do is to take a sequence in$A+B$, then ... 2 That does not quite work. Given a sequence$\{a_i+b_i\}$in$A+B$, if$\{n_i\}$and$\{m_i\}$are two sub-sequences of indices so that$a_{n_i}$converges and$b_{m_i}$converges, we can't just add these sequences because we want a subsequence of$\{a_i+b_i\}$.$\{a_{n_i}+b_{m_i}\}$is not in general a subsequence of$\{a_i+b_i\}$. We need$m_i=n_i$. The ... 2 Indeed, the other inclusion does not necessarily hold. Consider e.g. for fixed$n$the function$f_n\colon \mathbb{R} \to \mathbb{R}$given by$f_n(x) = \frac{1}{n} + \frac{1}{1+x^2}$. Then$\{ x\in \mathbb{R} : \lvert f_n(x)\rvert \leqslant 1/n\} = \varnothing$, but$\{ p \in \beta\mathbb{R} : \lvert\beta f_n(p)\rvert \leqslant 1/n\} \neq \varnothing$. ... 2 If$X$is non-empty, then there is no dependence on the axiom of choice. To see this, note that$X_\alpha$is a continuous map of$X$with the projection map$\pi=\pi_\alpha(x)=x_\alpha$. This follows from the fact that if a product$X=\prod_{i\in I}X_i$is non-empty, then for each$x\in X_i$there is a function$f\in X$with$f(i)=x$. Simply pick one ... 2 The term space is often used for the set with structure, so compact set/space is somewhat equivalent. Yet, there may be some subtle difference in being formal. A topological space is a pair$(X,\tau)$where$\tau$is a topology on a set$X$, even though we often just write$X$with$\tau$omitted. You can say that a topological space is compact (since its ... 2 Are you sure it holds if$U$is open? You can take the function$f$to just take everything to zero. And$X=[0,1]$and$U=(0,1)$. Then, I don't think$K$is compact. Here is a try if$U$is closed. Since the product of two compact spaces is compact, we have that the product space is compact. Thus, it suffices to prove that$K$is closed. Let$(x_n,t_n) ...

2

Recall that provability predicate are not "really" what you have expect them to be, just almost. It really just says that there is a number which encodes a proof sequence from statements which satisfy the condition "axiom of $T$" using particular inferences rules. This extends to non-standard models as well, only now the condition "axiom of $T$" as well the ...

2

Indeed, the sequence defined by $x_n = n$ works. To see this, let $y_n = x_{\phi(n)}$ be a subsequence. Let $y\in\mathbb R$. We have \begin{align} \lim_{n\rightarrow\infty} \frac{|y_n-y|}{1+|y_n-y|} &= \lim_{n\rightarrow\infty} \frac{|\phi(n)-y|}{1+|\phi(n)-y|}\\ &= \lim_{n\rightarrow\infty} ...

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