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7

One of my favorite textbooks is Klaus Janich's Topology, and he has a nice motivation for compactness I feel, namely why we should care about. This is in addition to my comment about compact subsets of a Hausdorff space being essentially like finite point sets. But he writes: In compact spaces, the following generalization from "local" to "global" ...


7

Maybe a disappointingly boring example: take $X = [0, \omega_1) \times \{0,1\}$ with the product topology, i.e. the disjoint sum of two copies of $[0, \omega_1)$. Taking as given the fact that the Stone-Cech compactification (SCc) of $[0, \omega_1)$ is $[0, \omega_1]$, I claim the SCc of $X$ is $Y = [0, \omega_1] \times \{0,1\}$. Clearly $|Y \setminus X| ...


6

The reason is that the intersection of infinitely many open sets need not be open: you need the set of $V_{q_k}$ to be finite in order to ensure that $V$ is actually a neighborhood of $p$, rather than merely some set containing $p$.


6

You are done, just need to clarify your notation. You have $\|W\|_{HS}^2=\sum_{n=1}^{\infty}\|We_n\|^2= \sum_{j=1}^{\infty}\frac{1}{j^2} < \infty$.


5

Consider the following as subspaces of $\mathbb R$ $\{0,1\}$ or in fact any finite set is compact and discrete $[0,1]$ is compact but not discrete. $\mathbb Z$ and $\left\{\frac1n:n\in\mathbb N\,\right\}$ are discrete but not compact. (But $\left\{\frac1n:n\in\mathbb N\,\right\}\cup\{0\}$ is compact and not discrete) $(0,1)$ and $\mathbb Q$ are neither ...


4

$\mathbb{Q}$ has a base of clopen sets and is zero-dimensional but is also a standard example of a non-locally compact space. Similarly, $\mathbb{Q}\times [0,1]^n$ has Lebesgue covering dimension $n$ but is not locally compact.


4

It's not clear at all what is $W$. And is $U$ an open set, or a collection or open sets? It says one thing, but it seems to treat it as the other. Finally, and most importantly, you didn't prove that every open cover of $A$ has a finite subcover. For this you need to take an arbitrary open cover, and produce a finite subcover. One good way to do that is to ...


4

The continuous image of a compact metric space need not even be first-countable. Consider $X = [0,1]$ with the usual topology, and $Y = [0,1]$ with the co-finite topology. The identity function $f : X \to Y$ ($x \mapsto x$) is clearly a continuous function. However $Y$ is not first-countable, let alone second-countable.


3

Not sure what $E$ is doing there, but here is how the proof usually goes : The map $$ \tau : f \mapsto (f(x))_{x\in X} $$ defines a continuous injection from the unit ball of $B'$ to $U$. Let $A$ denote the image of $\tau$, so it suffices to prove that $A$ is closed in $U$. Suppose $y \in U$ such that $\tau(f_{\alpha}) \to y$. Then, for every $x \in B$ ...


3

In Gillman & Jerison's Rings of continuous functions I found the following note 8.21 N.B. A number of authors have fallen into the trap of assuming then every countable, closed, discrete subset of a completely regular space is $C^\ast$-embedded. We have just seen a counterexample: [...] It seems likely that one of these authors, or someone ...


3

1) We have that every model $M$ of $T$ (i.e. $M \vDash T$) satisfy at least one $F_i$. 2) Consider the set of formulae $\{ \lnot F_i \}_{i∈I}$; no model of $T$ can satisfy it, becuase for each model $M$ of $T$ there is at least one $\lnot F_i$ that is not satisfied by it. Thus, applying Compactness theorem to the set of models of $T$, we have that exists ...


3

We can show the stronger (because we don't need that $G$ is Hausdorf, we don't need that $G$ is compact, and we can show that $V$ can be picked as symmetric neighbourhood of $1$) result Proposition. Let $G$ be a topological group and $A,B\subseteq G$ compact subsets. Then there exists a nonempty open set $V\subset G$ such that $1\in V$ and $V=V^{-1}$ and ...


2

All the open covers are finite, so they are themselves finite subcovers. Note, you can see every open cover is finite without writing them all out. If $\mathcal{U}$ is an open cover of $X$, $\mathcal{U} \subseteq \tau$; as $\tau$ is finite, $\mathcal{U}$ is finite.


2

Here is one way I like to think of it: Suppose you're trying to cover an infinite compact set, and you really want to give it an infinite cover that doesn't have a finite subcover. So you get a collection of infinite sets that looks like it nearly covers everything - maybe you've left behind a countable subset of an uncountable set or something. You must ...


2

Your examples: The discrete metric space on a finite set is compact. Closed bounded sets in $\mathbb{R}^n$ are compact. The discrete metric space on an infinite set is not compact. Many examples in $\mathbb{R}^n$ are available here, but open balls are probably the most easily visualized. If your goal is to study metric spaces rather than topological ...


2

The map $X \mapsto X^tX$ is a continuous function from the space of all $n \times n$ matrices to itself. Therefore the inverse image of the closed set $\{I\}$ is closed.


2

Recall that $x$ is a limit point of $A$ if and only if every neighborhood of $x$ meets $A$ on an infinite set. So in any case a discrete space will never be Weierstrass compact. Simply because $\{x\}$ is a neighborhood of $x$. So no infinite set has any limit point. So the only way a discrete space can be Weierstrass is that it is finite, and the ...


2

There exists $r = r(U,K)>0$ such that the closed disk $D_r(z)$ of radius $r$ centered at $z$ is contained in $U$ for all $z \in K$. By the mean value property (area version) and triangle inequality for integrals $$ |f(z)| = \frac{1}{\pi r^2}\left|\iint_{|w-z| \le r} f(w) \, du \, dv\right| \le \frac{1}{\pi r^2}\iint_{|w-z| \le r} |f(w)| \, du \, dv $$ ...


2

For example, put $A:=\{1,\frac{1}{2},\frac{1}{4},...\}$. Then the set $B:=\{0\}\cup\bigcup_n \left(\frac{1}{2^n} + \frac{1}{2^n} A\right)$ is compact (closed and bounded) and has derived set $B'=\{0\}\cup \{\frac{1}{2},\frac{1}{4},...\}$.


2

This is only a partial answer, but it's too long for a comment, so I'll post this as an answer and hope that nobody complains. For simplicity, let's call your presheaf $\mathcal{F}$. First, let's look at the stalk: The stalk $\mathcal{F}_x$ is just going to be the direct limit (with respect to the restriction you defined) of the spaces $U^*$ with $U$ an ...


2

The open interval $(0,1)$ (with the standard metric) is totally bounded but not compact (for example the intervals $(1/n,1)$ are an open cover with no finite subcover). The problem with the proof is that there's simply no reason each element of $U$ should be contained in some $S_n$. In the proof that totally bounded plus complete implies compact we first ...


2

In general, as Asaf pointed out, there are several spots that aren't very clear. That being said, I'll try my best to read between the lines and comment on claims I think you're making, and spots that I suspect you're confused. For a compact set $B$, let $A \subset B$ be closed. Thus for each open cover of $A$, we have $A \subset \cup_{i=1}^{n} G_i$ for ...


2

Hint 1: Consider the identity map $i : X \to X$ as a map between the metric spaces $(X, d_1)$ and $(X, d_2)$. Then $i$ is a homemorphism, i.e. $i, i^{-1}$ are continuous. Hint 2: A continuous function on a compact metric space is uniformly continuous. If you unwind the proof of Hint 2, you should be able to come up with a direct proof of your claim.


2

There is an order topology which satisfies this. Simply put two copies of $[0,\omega_1)$ back to back (so the space would look like $(−\omega_1,\omega_1)$. Formally, let $X=2\times \omega_1$ and put the lexicographic ordering on X with the usual ordering reversed in the first factor. So $(i,x)<(j,y)$ in $X$ if $i=0$ and $j=1$, or if $i=j=0$ and $x>y$, ...


2

Let $X$ be an arbitrary separable Tychonoff space (in particular, we can consider $X=\Bbb R$) and $n:\Bbb N\to X$ be an arbitrary map with dense image. Let $\beta n:\beta\Bbb N\to\beta X$ be the extension of the map $n$. Then $\beta n(\beta\Bbb N)$ is a compact dense subset of a compact space $\beta X$, so $\beta n(\Bbb N)= \beta X$. Since $\beta n$ is a ...


2

As @BrianM.Scott noted, every space $X$ has a finite open cover, namely $\{X\}$. So this tells us nothing about $X$. One way to motivate the definition is to see that if $X = \mathbb{R}^d$ with the standard topology, it is satisfied by precisely the subsets which are closed and bounded. To illustrate this, I'll show this equivalence in one direction: if ...


2

Let $x$ be a point in $X\setminus f^{-1}(A)$, and let $a_x$ be the closest point to $x$ from $A$ (such a point exists by compactness of $A$). Since $d(x,k_x)<d(x,a_x)$ by the uniqueness of $k_x$, we can consider the open ball $B_ε(x)$, where $$2\varepsilon=d(x,a_x)-d(x,k_x)>0$$ Take any $y\in B_ε(x)$. Can you finish from here and show that $y\notin ...


2

Chival mentioned in the comments that a normed space cannot be bounded (since it is a vector space). Rephrasing the question to overcome this issue, we get Let $X$ and $Y$ be finite dimensional normed spaces. Let $D:X \to Y$ be an isometric isomorphism. Prove that if $Z \subset X$ is compact then $D(Z)$ is also compact. Then, by noticing $D$ is ...


2

As others have noted, the term that you want is anticompact. This is an example of the kind of property studied by Paul Bankston in The total negation of a topological property, Illinois J. of Math., Vol. 23, Nr. 2 (1979), 241-252. I quote: Let $K$ be a topological class. The spectrum $\operatorname{Spec}(K)$ of $K$ is the class of cardinal numbers ...


1

Since $H_i$ is continuous we know that $H_i^{-1}(ClX_0)$ is a closed set which contains X. Since $Y_i$ is the closure of X, we know that $Y_i \subset H_i^{-1}(ClX_0)$. Therefore $H_i(Y_i) \subset ClX_0$.



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