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14

Yes. Let $X$ be any non-empty set. Perhaps the easiest way is to fix a point $p\in X$, let $Y=X\setminus\{p\}$, and set $$\tau=\wp(Y)\cup\{X\setminus F:F\subseteq Y\text{ and }F\text{ is finite}\}\;.$$ Then $\tau$ is a compact Hausdorff topology on $X$. It’s easy to check that it’s Hausdorff. If $\mathscr{U}$ is any open cover of $X$, there is a ...


6

If I choose $(0,1)$ and do a covering like this: $(-1,\frac{1}{2})$ and $(\frac{1}{3},2)$ it's a finite covering, then $(0,1)$ is a compact set? You are misunderstanding the definition of a compact set. The definition starts with "For every open coverings, there exists a sub coverings......". So, in order to show that $(0,1)$ is compact, you first have ...


5

It is compact. Your error is in thinking that a compact, totally disconnected space must be finite: this is false. A compact discrete space must be finite, but a totally disconnected space need not be discrete, as the present example shows. An even better example is the Cantor set, which is totally disconnected but, unlike this space, does not even have any ...


5

Consider $f:\mathbb R\to\mathbb R $ defined by $f (x)=1$. The set $\{1\} $ is compact, but $f^{-1}(\{1\})=\mathbb R $ is not.


4

A subset $A \subseteq \mathbb{R}$ under the standard topology is compact if and only if it is closed and bounded. Since the set $S$ is equal to the points of the sequence $(x_n)=(1/2^n)$ together with its limit $0$, it is closed. It is also clearly bounded and therefore compact.


4

The OP's own answer, the one-point compactification of an uncountable discrete space, is a classical one. Another is $\{0,1\}^X$, where $X$ is uncountable, where the set of all points with exactly one $1$ is uncountable and discrete (and so not Lindelöf too). Of course, every closed subspace of a Lindelöf space is also Lindelöf, so by the same argument, ...


4

The flaw with your proof is that you showed that one particular open cover has a finite subcover. To show compactness, you have to show that every open cover has a finite subcover. Suppose $A_n=\{x\in\mathbb R^n:1/n<\Vert x\Vert<2\}$. This is an open cover with no finite subcover, so $D^n\setminus\{0\}$ is not compact. Alternatively, you can see that ...


4

Hint. Let $U_n := X \setminus X_n$. If we have $\bigcap_n X_n = \emptyset$, we would have $\bigcup_n U_n = X$ with open $U_n$. Now use compactness.


4

Both sequential and covering compactness are purely topological notions. Moreover, if the metric $d$ induces the topology $\tau$ then there is no confusion.


3

$(0,1)$ is not compact. Proof: Consider the open covering $\displaystyle\left\{\left ({1\over n}, 1 - {1\over n}\right) \right\}_{n=3}^\infty$; this is an open covering of $(0,1)$ but it has no finite subcovering.


3

You write some $\epsilon$ as close as possible to zero But there is no such $\epsilon$! For any positive $\epsilon$, there is a strictly smaller positive $\epsilon$ - for instance, $\epsilon/2$. And the interval $({\epsilon/2}, 1)$ will contain some points in $(0, 1)$ that $(\epsilon, 1)$ does not (for instance, $\epsilon$ itself). While no specific ...


3

No. If $X=[0,1]$, then the sequence $\{e^{2\pi inx}\}_{n\in\mathbb{Z}}$ has no convergent subsequence.


3

That proof relies on a lemma. Lemma. Suppose $\mathscr{B}$ is a basis for the topology $\tau$ on $X$. Then $X$ is compact if and only if every open cover consisting of elements of $\mathscr{B}$ has a finite subcover. A basis for $\tau$ is a subset $\mathscr{B}$ of $\tau$ such that every open set $U\in\tau$ is the union of elements in $\mathscr{B}$. One ...


3

Unfortunately, this is not true. Take for example, any uncountable set $X$ with the discrete topology (so that singletons are open). It is obviously not Lindelöf. Now, 'create' the space $Y$ by adding a point $*$ to $X$ with neighbourhoods of the form $U=\{*\}\cup C$, where $C$ is a co-finite subset of $X$. To be clear, $Y=X\cup\{*\}$, with all points in ...


2

In order to carry out a one-point compactification, you start with a topological space. So you have to decide what topology you're going to give $[-\infty, \infty]$ and $[0, \infty]$. One possible topology you could give $[0, \infty]$ is to give a subbasis consisting of stuff like $[0,a)$ and $(a,\infty]$. That would give you the normal topology when you ...


2

Your arguments for uncountable and not dense look good to me. As far as showing $[0,1]-E$ is open, I think for simplicity since you know that $d_N\not\in\{4,7\}$, and we're dealing with integers, take $$\delta<\frac{1}{10^{N+2}}.$$ Then if $y$ is such that $|x-y|<\delta$ you know that $y$ has to agree with $x$ at $d_N$, and thus $y\not \in E$. I ...


2

HINT: $\mathscr{U}=\{U(0,p):p\text{ is prime}\}$ covers $\Bbb Z\setminus\{-1,1\}$. (Why?) Find open sets $V$ and $W$ containing $-1$ and $1$, respectively, such that $\mathscr{U}\cup\{V,W\}$ has no finite subcover. You may find Dirichlet's theorem useful.


2

Yes, it does. Since $W^{1,p}(U)$ is reflexive, there exists a further subsequence $\{u_{k_{j_i}}\}$ which converges weakly to some $v\in W^{1,p}(U)$. It must hence also converge weakly to $v$ in $L^p$, and since $u_{k_{j_i}}\to u$ in $L^p$ it follows that $u=v$ and hence has a weak derivative.


2

Since $X$ is compact, the image in $\mathbf{R}$ is compact (as $f$ is continuous). Let $y$ be the maximum of the image of $f$. Then $f^{-1}(y)$ is a closed set in a compact space, hence compact.


2

Begin with a fishing line of arbitrary but finite length. Proceed to cover it with a collection of pancakes whose edges have been removed. Observe that it takes a finite number of pancakes to cover the fishing line. Eat the pancakes. Enjoy math.


2

Consider say $[0,1] \times [0,1].$ There are open sets in the product which are not products of one open set from each factor, e.g. the interior of a circle contained in that square. I think by saying the open sets are generated by those you mention one is working with a subbasis for the proof, which seems like it is OK to do, but not directly without some ...


2

This part serves no purpose: Consequently for all $(x,y)\in X\times Y$ we have, $$B_{d_{X\times Y}}\left((x,y),\varepsilon_{(x,y)}\right)\subseteq\displaystyle\bigcup_{\gamma\in\Lambda} W_\gamma$$Hence, \begin{align*}\displaystyle\bigcup_{(x,y)\in X\times Y}B_{d_{X\times ...


2

A compact space is one in which every open cover has a finite subcover. Not every subcover has to be finite. So, for example, the cover $C = \{(-\varepsilon,\varepsilon)\,|\, 0 < \varepsilon\}$ is an open cover of $[-1,1]$, and has as a finite subcover the singleton $\{(-2,2)\}$. The cover is still infinite, however.


2

Theorem: If $X$ is countably compact and sequential (without any additional separation property and thus not necessarily $T_1$) then $X$ is sequentially compact. I found a proof in T. P. Kremsater, "Sequential Space Methods" (Master of Arts thesis). In the $T_1$ case the singleton sets $\{ x \}$ are closed. In the non-$T_1$ case consider instead the ...


2

Take the sequence $\large a_n = \frac{1}{n}$, which lives in (0,1). it converges to a point, $0$, which is not in your set (0,1). and so we say that your set is not sequentially compact. (and so it is not compact.)


2

Take a cover by open sets $U_\alpha$. First fix $x \in X$. The set $\{x\} \times Y$ is homeomorphic to $Y$ and so compact. It is covered by the $U_\alpha$ and so there is a finite subcover $U^x_{\alpha_1}, \dots, U^x_{\alpha_{n_x}}$ of $\{x\} \times Y$. Let $U_x := \bigcup^{n_x}_{i=1} U^x_{\alpha_i}$. Since $U_x$ is open, for each $y \in Y$ there is a ...


2

You can't just choose any such sets $U_x$; you have to choose them carefully such that no finite subcollection of them can cover $\mathcal{A}$. (Also, you don't need to choose $U_x$ for every $x\in X$, only for every $x\in\mathcal{A}$.) In fact, you can choose $U_x$ for each $x\in\mathcal{A}$ such that $U_x\cap\mathcal{A}=\{x\}$ for each $x$, and then it ...


2

Hint: Note that $|f|\leq 1$ (a better bound may exist, but this one is obvious). For $c>0$, write $$D_c=\{(x,y)\in\mathbb R^2 :f(x,y) \geq c\} = f^{-1}([c,1])$$ Since $(0,y)\notin D_c$, and $D_c$ is compact, we can find a "tube" $(-\epsilon,\epsilon)\times \mathbb R$ which doesn't meet $D_c$. Can you see that $D_+$ and $D_-$ are both compact now, where ...


2

Every connected component of a compact metric space is compact. Proof: Connected components are closed, and a closed subset of a compact space is compact.


2

Presumably the author assumed that all readers are familiar with diagonal sequence constructions and therefore just gave a broad outline, skipping some important details of the procedure. You don't choose the subsequences independently. One uses the diagonal sequence construction to make sure that the selected subsequence (the diagonal sequence) is a ...



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