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6

Ascending chains like that will not preserve compactness, since the finite subcover for each level might get bigger and bigger. As an example, look at $[-1,1] \subset [-2,2] \subset [-3,3] \dots$ in $\mathbb{R}$. Each set is compact, but the union is not.


6

Hint: look at infinite powers $X^I$ of finite discrete spaces $X$, e.g. $X=\{0,1\}$. This provides non-trivial Hausdorff examples. To get a metric example you need to take a little bit of care on (the cardinality of) the index set. Edit: $X$ need not be finite but can be any compact metric space.


5

For the first topology :- The only compact sets are finite sets. Clearly finite sets are compact so we just need to prove that any infinite set is not compact. Let $A$ be any infinite set. Consider the open cover $\{0,1,a\}_{a\in A}$. Then since $A$ is infinite this open cover does not admit any finite subcover and hence $A$ is not compact. Hint for the ...


4

We have the following result in Set Topology: Prop. Let $X$ be a space with the two properties that: (i) each point has a neighborhood basis of closed neighborhoods ($X$ is regular), and (ii) every continuous image of it in a Hausdorff space is closed. Then $X$ is compact. Proof. Suppose first $X$ is Hausdorff (the case of metric spaces). Let $\{U_i\}$ an ...


4

The product topology is the topology of pointwise convergence; your functions converge pointwise to the zero function. You are correct that $[0, 1]^{\mathbb{R}}$ is not sequentially compact; I believe $f_n(x) = |\sin nx|$ is an explicit counterexample, but I haven't checked it carefully. But neither sequential compactness nor compactness imply the other in ...


4

By construction, $(\mathbb{N}, d)$ is isometric to the subspace $\{\frac{1}{n} \mathrel{|} n > 0\} \cup \{0\}$ of $\mathbb{R}$ with the usual metric. This subspace is closed and bounded and hence compact by the Heine-Borel theorem.


4

Let $a,b$ be two elements of $M$. We wish to show that $d(f(a),f(b)) \leq d(a,b)$. As noted in the OP, there is an increasing sequence $(r_n)$ of integers such that $f^{r_n}(a)$ converges in $M$ and $f^{r_n}(b)$ converges in $M$. Then $d(f^{r_n}(a),f^{r_{n+1}}(a))$ tends to zero. But $d(a,f^{r_{n+1}-r_n}(a)) \leq d(f^{r_n}(a),f^{r_{n+1}}(a))$ by the ...


4

Trivially, no. Let $K=\emptyset$. The empty set is closed as $\mathbb{R}^n$ is open, certainly bounded. And given any two points $x,y \in \emptyset$, there is a path connecting them. :)


4

Here's a non-trivial example: the closure of the topologist's sine curve with the ends joined up. If you want an explicit representation, take $$ \{(x,y) : 0< x\leq 1, y = \sin(1/x) \} \bigcup \{(0,y): -1\leq y\leq 1\} \bigcup \{(x,0): -1\leq x\leq 0\} \bigcup \{(-1,y): -2\leq y\leq 0 \} \bigcup\{(x,-2): -1\leq x \leq 1\} \bigcup \{(1,y): -2\leq y\leq ...


4

Contrary to what I tried to do first, I think I now have a (complete) argument that your claim is false. Let me rewrite your last assumption (the displayed equation). Since $\int x^2|\widehat{g}(x)|^2 = \int |g'|^2$, this really says that certain derivatives are uniformly bounded in $L^2$ (originally, I may have to define these as distributional ...


3

I guess that $x\mapsto \frac{1}{2} x$ from $[0,1]$ into intself is injective, but is not an isometry. The answer is no.


3

If you want a finite interval example: $$\;\left[\frac12,\,1\right]\subset\left[\frac13,\,1\right]\subset\ldots\subset\left[\frac1n,\,1\right]\subset\ldots$$ Each interval is closed and bounded and thus compact, yet their union is not: $$\bigcup_{n=2}^\infty\left[\frac1n,\,1\right]=(0,1]$$


3

Products of compact sets are in fact compact. This is the Tychonoff theorem. As for unions, recall that singletons are compact, and every set is a union of singletons. What fails in your suggested proof is that it might be the case that each $x_n$ lies in a different compact set. In general, $X\subseteq A\cup B$ does not mean that $X\subseteq A$ or ...


3

Take the sequence {$ 1,2,3,...$} in $\mathbb R$ . Does it have any convergent subsequence?


3

No, it cannot. For any $x_1, \ldots, x_n \in X$, the linear span $V$ of $x_1, \ldots, x_n$ is finite-dimensional, so it is not all of $X$, and by Hahn-Banach there is a nonzero continuous linear functional $f$ such that $f = 0$ on $V$. Take $y \in X$ with $0 < \|y\| < r_0$ so that $f(y) > r \|f\|$. Now $$\|y - x_j\| \ge \|f\|^{-1} |f(y - x_j)| = ...


3

Yes it is, because it's sequentially compact: Let $x_n$ be a sequence in $\mathbb{N}$. Case 1: it takes on finitely many values. Then there is a constant subsequence which obviously has a limit in $\mathbb{N}$. Case 2: it takes on infinitely many values. Then it has a strictly increasing(in the usual sense) subsequence. This subsequence has to converge to ...


3

$X=\{0\}$ works, doesn't it? Both $X$ and $X\times X$ have only one element.


3

The arcs are also compact and connected, so their images must also be closed, bounded intervals, and they necessarily contain $\{a, b\}$.


2

It is not bounded, unless you are working with $1\times 1$ matrices. The only complex, $1\times 1$ orthogonal matrix are $(1)$ and $(-1)$. In $2\times 2$ (and by extension you get $n\times n$) you can consider the family, where $\lambda \geq 1$ is a real parameter, $$ A_\lambda = \begin{pmatrix} \lambda & i \sqrt{\lambda^2 - 1} \\ - i \sqrt{\lambda^2 ...


2

Hint: $\{0,1,2\}$ is open. The topology looks a lot like the discrete topology on $\Bbb R \setminus \{0,1\}$


2

We have to show that $\alpha (G)$ is closed. It seems the following. It is well known and easy to show the next Lemma. [Eng, Theorem 3.3.9] Every locally compact subspace $M$ of a Hausdorff space $X$ is an open subset in the closure $\overline{M}$ of the set $M$ in the space $X$. Now, since the space $\alpha (G)$ is locally compact, it is an open ...


2

This is tied in to a standard set of compactness results for metric spaces. Here the usual sequence of equivalent properties is as follows: Countable compactness Limit point compactness (every infinite set has a limit point) Sequential compactness (every sequence has a convergent subsequence) Compactness These are all equivalent for metric spaces. I'm ...


2

Any one-point space (which is finite, yes) clearly verifies this... And now, you're going to edit your question and say : "please help to find a compact $X$ that is not one point, such that $X$ and $X\times X$ are homeomorphic"...


2

Assume the space is covered with open sets $U_i$, $i\in I$. Then $0$ must be in one of these sets, so there must be one open set V among the $U_i$ cotaining $0$. So we can find a natural number $m$ s.t $0 \in B(0, 1/m) \subset V$ where $B$ is the open ball in this space. Now the numbers $\{m+1, m+2, \dots\}$ are in this ball. Take $V$ and the open sets ...


2

You are right, but it could be simplified a bit more. $A$ is an open ball of radius $\frac{1}{2}$ centered at $\frac{1}{2}$. So it is bounded. It is an open set, therefore it is not compact. An explicit example for an open cover that does not have a finite subcover would be the balls centered at $x_n$ with radius $r_n$, where $x_n$ is $\frac{1}{n}$ and ...


2

To start, notice that the intersection of any chain of nonempty compact sets must be nonempty (by the finite intersection property). Let $D = \cap \mathfrak{C}$. Then $D$ is nonempty by the preceding statement. To prove that $D$ is compact, choose some $C_0 \in \mathfrak{C}$. Since $X$ is Hausdorff, all compact subsets of $X$ are closed in $X$, and thus all ...


2

It seems the following. Fix any element $K\in\mathfrak{C}$. Then the set $C=\bigcap \mathfrak{C}$ is compact as an intersection $\bigcap\{ K\cap L:L\in\mathfrak{C}\}$ of a family $ \{ K\cap L:L\in\mathfrak{C}\}$ of closed subsets of a compact set $K$. Assume that the set $C$ is not connected. This means that $C$ can be represented as a union $C_1\cup ...


2

There is a sequence $x_n \in K$ such that $|x_n - p| \to\ \mathrm{inf} \{|x - p| : x \in K \}$. As $K$ is compact, every Cauchy sequence in $K$ converges within $K$. So $x_n$ converges to some $x$. Similarly for the supremum. It does not work for closed sets, because for example the supremum can be infinite.


2

Compact Lie groups are topological groups whose topology is compact. A compact Lie algebra is the Lie algebra of a compact Lie group, therefore the name "compact". Lie algebras are vector spaces, and all finite-dimensional Lie algebras are linear, i.e., subalgebra of the Lie algebra of matrices. For Lie algebras "compact" just means that they are reductive ...


1

Hint 1: what do you know about the sequence $(f(n))_{n=1}^{\infty}$? Hint 2: note it says the definition of sequential compactness. Hint 3: Recall the definition of convergence: a sequence $(f(n))$ converges to $\alpha$ if for any given $\varepsilon>0$, there is an $N$ such that $$ \left| f(n) - \alpha \right| < \varepsilon. $$ So, we have a ...



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