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10

Yes. Since $\emptyset$ is compact, $X$ is compact, so every closed subset of $X$ is compact, so every open subset of $X$ is compact. If every open subset of $X$ is compact, then every subset of $X$ is compact. That is, every family of open sets in $X$ has a finite subfamily with the same union.


9

For $n\in\Bbb Z^+$ let $x^{(n)}=\langle x_k^{(n)}:k\in\Bbb Z^+\rangle$ be the sequence defined by $$x_k^{(n)}=\begin{cases} 1,&\text{if }k=n\\ 0,&\text{otherwise}\;. \end{cases}$$ Clearly $\langle x^{(n)}:n\in\Bbb Z^+\rangle$ is a sequence in $A$, and it has no convergent subsequence; indeed, any two distinct terms are distance $1$ apart.


5

Let our Banach space be $\ell^1$, which is the set of all $a \in \mathbb{R}^{\mathbb{N}}$ such that $\sum_n |a_n|<\infty$. The norm is the $L^1$ norm. For $i \in \mathbb{N}$, let $e_i \in \ell^1$ denote the $i^{th}$ standard basis vector, i.e, the sequence whose $i^{th}$ index is $1$, but all other indices are $0$. Consider the linear map $T: \ell^1 \to ...


5

For the first part, you can use an open cover $\{B_{N}(0)\}$ of balls of increasing size. Since the set is unbounded, there is no finite subcover. For the second part, let $x \in \overline{S}\setminus S$. You can then use the open cover $\{ \Bbb R^N \setminus \overline{B_{1/n}(x)}\}$ of complements of closed balls. This is a cover since $x\not \in S$, so ...


4

Homeomorphisms, in general, preserve neither completeness nor total boundedness. Consider $\Bbb R$ and $(0,1)$.


4

No. The general characterization is that a metric space is compact if and only if it is complete and totally bounded. The latter means that for any $\varepsilon > 0$ the space has a finite cover by balls of radius at most $\varepsilon$ (this is sometimes called an "$\varepsilon$-net"). This rules out, for instance, the closed unit ball in an infinite ...


3

No. Consider $(\mathbb R,d)$ by the metric: $d(x,y)=\frac{|x-y|}{1+|x-y|}$.


3

You definitely need the target space to be Hausdorff. Theorem $\mathbf{5.4}$ of Gerlits, Juhász, Soukup, & Szentmiklóssy, Characterizing continuity by preserving compactness and connectedness, says that if $X$ is $T_3$, and every connectedness-preserving, compactness-preserving map from $X$ to a $T_1$ space is continuous, then $X$ is discrete. Thus, ...


3

My favourite proof is due to Hartig: D. G. Hartig, The Riesz representation theorem revisited, American Mathematical Monthly, 90(4), 277–280. Hartig claims that his proof is ...category-theoretic but having unwrapped the details I must say it is really functional-analytic. The idea is as follows. Step 1. We can easily prove this theorem for extremely ...


2

It seems the following. Define a map $h:X\to Y\times Y$ by putting $h(x)=(f(x),g(x))$ for each point $x\in X$. Since both maps $f$ and $g$ are continuous then the map $h$ is continuous too. Since $X$ is a compact space, $h(X)$ is a closed subspace of a Hausdorff space $Y\times Y$. Moreover, since the set $h(X)$ does not intersect the diagonal ...


2

Given $t\in[0,1]$ there exists an open nhbhd $U_t$ of $t$ in $[0,1]$ such that for every $s\in U_t$ $A_s^{-1}A_t\in V$ (since $A_t^{-1}A_t=I\in V$ and $V$ is open). The collection of $U_t$'s give you an open cover of $[0,1]$. The "standard compactness argument" referred above is that since $[0,1]$ is compact you can select a finite subcover of $[0,1]$, ...


2

Yes, a morphism in this category is epic iff it is surjective. Obviously surjective morphisms are epic, so it suffices to show that if $f: X\to Y$ is not surjective, there is a locally compact Hausdorff space $Z$ and distinct $g_1, g_2 \in \hom(Y,Z)$ such that $g_1 \circ f = g_2 \circ f$. I will use the notion of a perfect map as defined in Engelking's ...


2

Hint: Recall that for metric spaces, compactness is equivalent to sequential compactness. Consider the sequence of sequences $(x_n)_m = 1$ if $n=m$ and $0$ if $n\not=m$. Every element has distance $1$ from the $0$ sequence. Can it have a convergent subsequence?


2

Hint: consider an open cover consisting of open balls of radius $1/3$ around every point of $B[x,1]$. Find infinitely many points whose pairwise distances are all $> 2/3$.


2

Just to cover both results: Assume that $(x_n)_{n\in\mathbb{N}}$ is a Cauchy sequence in a sequentially compact space $X$. Introduce a convergent subsequence $(x_{n_k})_{k\in\mathbb{N}}$ of $(x_n)$ with $x_{n_k}\to x.$ Let $\epsilon>0$ be given. Choose $N$ such that $\rho(x_i,x_j)<\epsilon/2,~i,j\geq N$. Choose $n_k>N$ such that ...


2

To me the intuitive notion of compactness is sequential compactness, i.e. that every sequence has a convergent subsequence, and the open cover definition is only motivated by the Heine-Borel theorem. Of course the open cover definition is more general, in that it makes sense in any topological space, but understanding what it really means takes quite a bit ...


2

A space $X$ is a $k$-space if it has the final topology with respect to all maps from compact Hausdorff spaces to it, in other words, if $A\subseteq X$ is closed if $t^{-1}(A)$ is closed in $K$ for every map $t:K\to X$ where $K$ is compact Hausdorff. Usually, a compactly generated space has the property that a subset is closed if it intersects every compact ...


2

Some quotations from “General Topology” by Ryszard Engelking:


2

We have $A\times B\subseteq N\subseteq X\times Y$, where $A$ and $B$ are compact. For each $(x,y)\in A\times B$, we know there is a basic open set $U_x\times V_y$ where $x\in U_x$ is open in $A$ and $y\in V_y$ is open in $Y$, and $(x,y)\in U_x\times V_y\subseteq N$. The product of compacts is compact, so $A\times B$ is compact. Thus $\exists$ a finite ...


2

Choose $\displaystyle Q_n = \bigg(\frac1n, 2\bigg)$ for $n = 2, 3, \ldots$. Notice that $\displaystyle (0, 1] \subseteq \bigcup_{k = 2}^\infty Q_n$ so the $Q_n$ form an open cover. Suppose, by way of contradiction, that $\{Q_n\}$ has a finite-subcover, say $Q_{k_1}, \ldots, Q_{k_n}$. Choose $k = \max\{k_1, \ldots, k_n\}$ and observe that $\displaystyle ...


2

The map $$ F: \mathbb{R}^{n\times n} \to \mathbb{R}^{n\times n},\, F(X)=XX^T $$ is continuous (it is actually differentiable), and $$ O(n)=\{A\in \mathbb{R}^{n\times n}:\, AA^\top=I_n\}=F^{-1}(I_n). $$ Since $\{I_n\}$ is closed and bounded in $\mathbb{R}^{n\times n}$, and $F$ is continuous therefore $O(n)$ is closed and bounded because each entry $a_{ij}$ ...


2

Your proof is wrong. When you write 1)Suppose K is disconnected. Then we write it's separation as K=U∪V. Hence K is open since U and V are open. you've gone off the rails. For $K$ is disconnected if there are open sets $U$ and $V$ in $K$'s topology with the property that you specify. But what does it mean for $U$ to be open in $K$? It means that $U$ ...


2

The problem here is that you misunderstand compactness. A set $S$ is compact if and only if, for ANY covering of $S$ with open sets, there exists a finite subcovering which also covers $S$ Yes, indeed, $\mathbb R$ does cover $\mathbb Q$, meaning that $\{\mathbb R\}$ (i.e. the SET that contains one element, the set of real numbers) is a covering of ...


2

$\newcommand{\Reals}{\mathbf{R}}$Comparing this question with your linked question, the central issue seems to be the term "closedness", which perhaps feels bothersome because manifolds are unions of open sets. If that's really the question, the resolution comes down to "relative topology", how "open" and "closed" are defined for subsets of $\Reals^{n}$. ...


2

Use functions that are $n^2$ on $[0,1/n]$ and zero on $[1/n+\epsilon,1]$ and on $[1/n,1/n+\epsilon]$ connect them to be continuous. Then the integral of any one of them is close to $n$ (choose an appropriate $\epsilon$ that will depend on $n$) and I believe the integral of the difference of any two of them is at least one.


2

You many want to assume that $N$ is Hausdorff in order to conclude that $f$ is a homeomorphism. To show that $f$ is homeomorphic, it suffices to show that it maps closed subset to closed subset (i.e. $f^{-1}$ is continuous as well). Any closed subset in $M$ is compact, so it get sent to a compact subset of $N$. If $N$ is Hausdorff, then this compact subset ...


2

Let $$X=\bigl((\mathbb R\setminus\mathbb Q)\times\{-1,1\}\bigr)\cup\bigl( \mathbb Q\times\{0\}\bigr) $$ where we declare $U\subseteq X$ open iff $U=(V\times\{-1,0,1\})\cap X$ for some open subset $V$ of $\mathbb R$. Admittedly, $X$ is not Hausdorff. But it contains two subsets $(\mathbb R\times\{0,1\})\cap X$ and $(\mathbb R\times\{0,-1\})\cap X$ that are ...


2

It suffices to show that: Let $U$ be an open set which contains $[a, b]$, then $m(U) > b-a$. (Let $U$ be the union of your open intervals in $F$, then the quantity you want is $\ge m(U)$. As $[a, b] \subset U$, we have $m(U) \ge b-a$. To show that we have strict equality, it suffices to show that $m(U\setminus [a, b])>0$. But as $U\setminus [a, b]$ ...


2

It seems the following. A topological group $G$ is called $\omega$-precompact or $\omega$-bounded, if for each neighborhood $U$ of the unit there exist a countable subset $C$ of the group $G$ such that $G=CU=UC$ (or, equivalently, if for each neighborhood $U$ of the unit there exist a countable subset $C$ of the group $G$ such that $G=CU$). It is obvious, ...


2

Counterexample. As in Alex Ravsky's example, let $V=\omega$, $E=\{(m,n):m\gt n\}$, $L=\{0,1\}.$ Now simply define $f(p_0,p_1,\dots)$ to take the value $1$ if all of the inputs are $0$, and to take the value $0$ otherwise.



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