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17

A compact set must be bounded. Otherwise we can take $\{ x_n \}_{n=1}^\infty$ such that $\| x_n \| \geq n$. This will have no convergent subsequence, which we can prove by showing that it has no Cauchy subsequence. A compact set must be closed. Otherwise we can pick a sequence which converges to a point in the closure which is not in the set. This will ...


7

They seem to be usually called KC-spaces (Kompact Closed), occasionally TB-spaces, and very rarely $J_1^\prime$-spaces. As you noticed, this class of spaces lies strictly between the T1-spaces and the Hausdorff spaces. I am unaware of any characterisation of them apart from the definition given. The closest thing of this kind I can think of is that a ...


5

I need a couple of preliminary results. Lemma $\mathbf 1$: Suppose that $X$ is $T_1$ and weak Hausdorff. Let $K$ be a compact Hausdorff spaces and $f:K\to X$ continuous; then $f[K]$ is not just compact, but also Hausdorff. Proof: Let $x$ and $y$ be distinct points of $f[K]$, and let $H_x=f^{-1}[\{x\}]$ and $H_y=f^{-1}[\{y\}]$; $H_x$ and $H_y$ are ...


4

Every compact metric space is separable, so let $D=\{x_k:k\in\Bbb N\}$ be a dense subset of $X_1$. For each $n\in\Bbb N$ let $D_n=\{x_k\in D:k\le n\}$; $D_n$ is finite, so there is an isometry $f_n:D_n\to X_2$. For each $k,n\in\Bbb N$ with $k\le n$ let $y_k^n=f_n(x_k)$, so that $f_n[D_n]=\{y_k^n:k\le n\}$. Note that for any $k,\ell,n\in\Bbb N$ with ...


4

The following result is available: Let $(X,\lVert\cdot\rVert)$ be a Banach space and $S$ a subset of $X$. Then $S$ is compact if and only if the following two conditions hold: $S$ is bounded; for each positive $\varepsilon$, there exists a finite dimensional space $F=F(\varepsilon)$ such that for all $x\in S$, we have $d(x,F)=\inf\{\lVert ...


3

Since $\operatorname{cl}_{\beta\omega}Y\subseteq\operatorname{cl}_{\beta\omega}X$, you might as well work in the subspace $\operatorname{cl}_{\beta\omega}X$. In that case you can assume without loss of generality that $X=\omega$. Fix $p\in(\operatorname{cl}_{\beta Y}Y)\setminus Y$; if we view $p$ as an ultrafilter on $\omega$, then $Y\in p$, and there is an ...


3

Not quite. A metric space space can be complete without being compact (e.g., $\mathbb R$ with the Euclidean topology). For a metric space, completeness + total boundedness = compactness.


3

Hint: Fix any element of an open cover of $K$ that contains $p_0$. Finitely many remain out there.


3

If $(X,\tau_X)$ and $(Y,\tau_Y)$ are homeomorphic, then there is a bijection between the two spaces. Namely, there is a function $f\colon X\to Y$ which is a bijection, and satisfies that $f[U]$ is open if and only if $U$ is open (for $U\subseteq X$). Now ask yourself, is there a bijection $f\colon\Bbb R\to\Bbb Z$?


3

If $X$ is metric and not compact, then $X$ has an infinite closed discrete subset $D$. By passing to a subset if necessary, we may assume that $D=\{x_n:n\in\Bbb N\}$ is countably infinite. Now define $f:D\to\Bbb R:x_n\mapsto n$, and apply the Tietze extension theorem to get an unbounded continuous real-valued function on $X$.


3

In order to prove that the space $\omega_1$ is locally compact, you have to prove that every point in $\omega_1$ has a compact neighborhood. If $\alpha\in\omega_1$ then the one-point set $\{\alpha\}$ is obviously compact, but it is not necessarily a neighborhood of $\alpha$; namely, it's not a neighborhood if $\alpha$ is a limit ordinal, such as $\omega$. ...


2

You can write each $z_n$ in that way. Since $A$ is compact, there exists a convergent subsequece $(x_{n_j})$. Now, the subsequence $(y_{n_j})$, also admits a convergent subsequence $y_{n_{j_k}}$. So the sequence $(z_n)$, admits a convergente subsequence $z_{n_{j_k}}$, and $AB$ is compact.


2

I think you are on a right way, and in your proof you can use the fact that there is no infinite decreasing sequence of ordinals.


2

You can explicitly write down a parametrization of $K$, showing that it's a continuous image of a compact set. But here is a general argument. Claim If $A\subset \mathbb R^n$ is compact and $C$ is the smallest convex set containing $A$, then $C$ is compact. Proof. The set $C$, also known as the convex hull of $C$ consists of all finite convex ...


2

Homeomorphic spaces share their topological properties. A discrete space with more than one element is not connected; the real line with the usual topology is connected. The compact subsets of a discrete space are finite, the real line has plenty of infinite compact subsets. Sequences with pairwise distinct terms are not convergent in a discrete space, ...


2

I think the only limit point of the first set $S_1$ but not in $S_1$ is $(0,0)$. Since for any $p=(u,v)\neq (0,0)$ and $p$ is not an element of the first set $S_1$, let $r^2=u^2+v^2$, we can find sufficient large $x$, s.t $e^{-x}<r$. Then we can find a neighborhood of $p$, which doesn't intersect $S_1$. Hence $p$ is not a limit point. Since $(0,0)$ is ...


2

An open cover of $(-2,2)$ can only be broken down into $(-2,0) \cup (0,2)$ if those two open sets are included in the cover, not to mention that $(-2,0) \cup (0,2)$ fails to cover $(-2,2)$ to begin with (what contains $0$?). You are right when you say that open sets of $\mathbb R$ are not compact (well the empty set is). To show this, you would have to ...


2

For instance, $f:[-1,1] \rightarrow \mathbb R$ defined as $f(x)=1/x$ if $x\neq 0$ and $f(0)=0$ is defined on a compact domain $[-1,1]$ but it is not bounded. Recall the Weierstrass theorem: "Every continuous function on a compact domain has at least one maximum and one minimum" So negating the above statement we obtain that: "No maximum or minimum and ...


1

consider for example $$f:[0,1]\to \Bbb R\\ f(x) = \begin{cases} \frac 1x &\text{if} & \frac 1x\in\Bbb N, x\neq 0 \\ 0 &\text{otherwise} \end{cases}$$ which is unbounded on $[0,1]$.


1

Generalization of completeness to non-metric spaces goes through the concept of uniform spaces.


1

No. Here is a counterexample, inspired by the answer to this question: If $V \subset H$ compact, is $L^2(0,T;V) \subset L^2(0,T;H)$ compact too? Take $v\in H^1$, $T=\pi$, $$ \phi_n(t) = \sin(n t)v. $$ Since $\sin(n \circ)$ converges weakly but not strongly to zero in $L^2(0,T)$, it follows that $\phi_n$ converges weakly to zero in $L^2(0,T;H^1)$. If ...


1

HINT: For any $p\in\Bbb R^2$ and any $\epsilon>0$, the set $\Bbb R^2\setminus\operatorname{cl}B(p,\epsilon)$ is open. (Here $B(p,\epsilon)$ is the open ball of radius $\epsilon$ and centre $p$.) Added: This doesn’t arise in your specific example, but in general you need more than that $p$ is in the boundary of the compact set: you need it to be a limit ...


1

Let $X$ be a compact Hausdorff space and $p\in X$ apoint such that $Y:=X-\{p\}$ is not closed. For each $x\in Y$ there exist disjoint open sets $x\in U_x,p\in V_x$. Then the $U_x$ cover $Y$. Assume there is a finite subcover $U_{x_1}\cup\ldots\cup U_{x_n}$. Then this subcover misses the open set $V_{x_1}\cap \ldots\cap V_{x_n}$ which contains $p$ ans must be ...


1

Let $p=(0,-1) \in \mathbb{R}^2$, and consider the collection of open balls $\{B(p,2-\frac{1}{n})\}_{n \in \mathbb{N}}$


1

Let's do it according to the most standard definition of compactness: A subset $K$ of a metric space $X$ is compact if every open cover has a finite subcover. Let $\{U_i\}_{i\in I}$ be an open cover of $K$. The $p_0\in U_{i_0}$, for some $i_0\in I$. But as $U_{i_0}$ is open, then a whole ball $B(p_0,\varepsilon)\subset U_{i_0}$. Next, as $p_n\to p_0$, ...


1

$K_1$ might have lots of points in common with other $K_\alpha$s, indeed it must since every finite sub-collection (such as every intersection of two sets) has nonempty intersection. The point is that we assume that no point of $K_1$ is in every $K_\alpha$ (by way of contradiction). I suppose you could assume (again by way of contradiction) that there was ...


1

For each $a\in\mathbb R$, let $K_a=[a,\infty)$. Note that $K_a$ is a family of closed subsets of $\mathbb R$ in which every finite subcollection has non-empty intersection, however $\bigcap K_a=\varnothing$. So you need compactness at some point.


1

I really don't like the way the proof is presented in Rudin, and I'd go as follows: suppose $\;\bigcap_\alpha K_\alpha=\emptyset\;$ , but then fixing $\;\alpha_0\;$ we get (de Morgan): $$K_{\alpha_0}=K_{\alpha_0}\setminus\emptyset=K_{\alpha_0}\setminus\bigcap_\alpha K_\alpha=\bigcup_\alpha\left(K_{\alpha_0}\setminus K_\alpha\right)$$ But ...


1

If the conclusion fails, then there is, in $X$, a first element $\beta$ such that $]\leftarrow,\beta]$ is not compact. Let $C$ be an open cover of $]\leftarrow,\beta]$ with no finite subcover, and let $U$ be an element of $C$ that contains $\beta$. Of course, $U$ doesn't include all of $]\leftarrow,\beta]$, so, by definition of the topology, it includes ...


1

Prove by induction on $\beta$ that $L(\beta) = \{x: x \le \beta \}$ is compact, for all ordinals $\beta$. This is clear for $\beta = 0$, where $L(0) = \{0\}$ and if $\beta+1$ is a successor, then $L(\beta+1) = L(\beta) \cup \{\beta+1\}$, so if $L(\beta)$ is compact, so is $L(\beta+1)$. So assume $L(\alpha)$ is compact for all $\alpha < \beta$ and ...



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