# Tag Info

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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