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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" ...

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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 ...

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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 ...

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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 ...

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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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