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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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