Tag Info

Hot answers tagged

6

postscript: I was a bit hasty. In a Hausdorff space, compact sets are closed. end of postscript Being closed [but see above], compact sets contain all of their boundary points. The boundary of a compact set is therefore a subset of that compact set and is closed. Closed subsets of compact sets are compact.


5

The positive integers work fine... They're closed because they contain no limit points (suppose one exists, derive obvious contradiction, etc) and they're unbounded by the archimedian property.


5

HINT: Your argument for (1) is correct. You can indeed use the same idea for (2). Note that (1) essentially proves the special case of (2) in which $A=\{a\}$ for some $a\in X$. In general, then, you can separate each $a\in A$ from $B$ with disjoint open sets. And then?


4

Your basic intuition is correct, and if $X$ were a metric space, or even a first countable space, you could show that $f$ is continuous by showing that it preserves limits of convergent sequences, starting off in just this way. Here, though, you know only that $X$ is Hausdorff; that means that its topology isn’t necessarily determined by its convergent ...


4

If $x \in (0,1)$ then $0<x<1$. By the Archimedean property of $\mathbb{R}$, we can conclude that that there exist natural numbers $m,n$ such that $1/m < x < 1-1/n$. In particular we can take $k=\max \{ m,n \}$ to get $1/k < x < 1-1/k$. I think you can conclude your result from here.


4

Idea: Show that $Q$ is sequentially compact using a diagonalization procedure. Let $(x^k)_{k \in \mathbb{N}} \subseteq Q$ be a sequence of the form $$x^k = \sum_{n=1}^{\infty} c_n^k u_n$$ with $|c_n^k| \leq \frac{1}{n}$. Since $(c_1^k)_{k \in \mathbb{N}} \subseteq [-1,1]$ and $[-1,1]$ is compact, we can pick a subsquence $k^1(j)$ such that $c_1^{k^1(j)} ...


3

The converse is not true. Consider the set of continuous functions $[0,1]\to \mathbb{R}$ in the box topology. $X$ is Hausdorff, and a sequence $x_1,x_2,\ldots$ of points of $X$ converges if and only if it is eventually constant by a diagonal argument using the fact that if two continuous functions differ at only finitely many points then they are equal. ...


3

To get your result, you have to choose correctly the function $f$; otherwise, you'll get nowhere. The easiest way to prove what you want is to argue by contradiction. So, assume that $X$ is not complete, and let $(x_n)_{n\in\mathbb N}$ be a Cauchy sequence in $X$ which is not convergent. Without loss of generality, we may assume that the $x_n$'s are all ...


3

In a chart around $p$, consider the set of all closed balls centered at $p$ of radius $1/n$ for some integer $n$. Every finite intersection of sets in this collection is nonempty but the intersection of all of them is empty. Since the manifold is Hausdorff and closed balls in the chart are compact, they are closed in the manifold. Once a point is removed, ...


3

Your idea from the previous thread was almost correct. HINT: Show that if $A\subseteq\Bbb R^n$ is closed and bounded, then it is a subset of some product of closed intervals (in fact, you can use the same interval!).


2

Let $B \subseteq \mathbb{R}$ be any bounded subset. Then "$B$ attains its bounds" means that $B$ is closed. Call $C(A)= \{ f: \mathbb{R}^n \longrightarrow \mathbb{R} \mbox{ continuous functions} \} $. Then $$A = \bigcap_{f \in C(A)} f^{-1}(f(A))$$ Since for all $f \in C(A)$ $f(A)$ is closed, $f^{-1}(f(A))$ is closed as well (because $f$ is continuous). ...


2

It is known that every compact $T_2$ space is normal. We will prove that if $X$ is infinite, then there is an infinite set of linearly independent real-valued continuous functions on $X$. Choose countably many points $\{x_n:n=0,1,2,\cdots\}$ in $X$ and choose an open neighborhood $U_n$ of $x_n$ which does not contain $x_i$ for all $i<n$. Since $X$ is ...


2

Here's an even easier way to show $\mathbb R^n$ has this for all $n\geq 2$, basically using your $S^1$ argument. The complement of $A\cup B$ is open and non-empty, so contains a pair of small open balls $U\subsetneq V$. Now take $C=U^c$. This is closed and contains $A\cup B$ by construction, and is (path) connected when $n\geq 2$.


2

This notion is called tightness of a sequence of measures. You can apply it in probability theory with the sequence $P\#{X_n}$ of the image probabilities under the action of $X_n$: $$ P\#{X_n}(A) = P(X_n\in A) $$ You transfer the topology issues to the (metric, often polish) space $\mathcal X$ where $$ X:\Omega \to \mathcal X $$ Note also that in the ...


2

Let $S=\{2^{-n}:n\in\Bbb N\}$, and let $K=\{0\}\cup S$. Let $X=\{0,1\}\times K$. Points of $\{0,1\}\times S$ are isolated. For $i\in\{0,1\}$ let $p_i=\langle i,0\rangle$, and let $\mathscr{U}$ be a free ultrafilter on $\Bbb N$. A set $V\subseteq X$ is a nbhd of $p_i$ iff $p_i\in V$, $(\{i\}\times S)\setminus V$ is finite, and $\{n\in\Bbb N:\langle ...


2

Let $X=\{(x,y)\in\mathbb R\times\mathbb R:x\ne0\text{ and }y=\sin\frac1x\}\cup\{(0,0)\} $ with its induced topology as a subspace of $\mathbb R\times\mathbb R$. Then $X$ is a connected (but not pathwise connected) metric space. There is no connected compact subspace of $X$ containing $\{(0,0)\}$ and any other point. If $x_1\lt0\lt x_2$, there is no connected ...


2

The Heine-Borel Theorem is valid only some "special kinds" of spaces, like a metric space as $\Bbb{R}$. If a metric space is not complete, then the theorem is not valid, for example take the space of the rational numbers (this is in fact a metric space, try prove is not hard) does not have the "Heine-Borel property".


2

In a general topological space you only have open sets to work with. Distance, and thus boundedness, does not necessarily make sense in that setting. Compact sets retain many of the intuitive topological and metric properties that finite sets have. The essence of it is contained in the finiteness of the cover.


2

It's not correct. Take $X_1 = (0,1)$ and $X_2 = \mathbb R$ (identified with $X_1$ via some homeomorphism). Then with the standard metric on both $X_i$, $X_1$ is totally bounded and $X_2$ is complete. Both are not compact.


2

Update: NOT true. Counterexample: Consider $\mathbb N$ with the discrete topology, which is clearly not compact. Let $d_1$ be the discrete metric, which is complete, and $d_2(n,m)\equiv|1/n-1/m|$. Now, $d_2$ can be shown to be a totally bounded metric. It is also not difficult to check that both $d_1$ and $d_2$ generate the discrete topology on $\mathbb N$. ...


2

Actually, $\{ x_A : A \in \mathcal{A} \}$ is always a zero-set in $\Psi(\mathcal{A})$: define the function $f : \Psi(\mathcal{A}) \to [0,1]$ by $$f(n) = \tfrac{1}{n+1} \\ f(x_A) = 0.$$ This is clearly a continuous function (any neighbourhood of $0$ contains all but finitely many of the $\frac{1}{n+1}$, and so it's inverse image under $f$ includes a cofinite ...


2

It takes a fair bit of work to write this out completely, and at least one idea that may not be very obvious. I’ll do most of the harder bits, but I’ll leave a reasonable amount of detail for you to fill in. You really would be better off working with the actual characterization stated in the theorem. Let’s start with a compact $K\subseteq\Bbb R^2$. In any ...


2

How to prove this depends a great deal on what tools you already have. The first answer assumes that you know that a metric metric space is compact if and only if it’s complete and totally bounded, but I shouldn’t be at all surprised if this exercise were intended to prepare for that result. Suppose that whenever $A$ is an infinite subset of $X$ and ...


2

Compactness is something relative to the topology. You are given a topology on $X$, namely $\tau=\{\varnothing,X\}$. And the claim is that every subset of $X$ is compact in that topology. So your counterexample is not a counterexample, since it doesn't uses the given topology.


2

If $V$ and $W$ are normed vector spaces of the same finite dimension $n$, then $V$ and $W$ are isomorphic as topological vector spaces, i.e., there is a a vector space isomorphism $f : V \rightarrow W$ such that both $f$ and $f^{-1}$ are continuous. However, if $n \ge 2$, $V$ and $W$ may not be isometrically equivalent, i.e., it may be impossible to choose ...


1

It depends on definition. It may be: $X$ is compact if each of its open covers has a finite subcover. or $X$ is compact if it is Hausdorff ($T_2$) and each of its open covers has a finite subcover. There is only one nonempty open set, hence $X$ is compact in the sense of 1. But even if $X=\{0,1\}$ with the topology in question, it is not Hausdorff, ...


1

A metric space $X$ is compact if and only if it is complete and totally bounded. If your $X$ is not compact, it can't be totally bounded. That means there's an infinite $\varepsilon$-packing $A$, which is exactly the set you need.


1

Let me organize things a bit for you so it's clear what you have to do: It's easy to see that $A$ is bounded $\iff$ $X(A)$ is bounded. Therefore you have prove only that $X(A)$ is closed $\iff$ $A$ is closed and $X(A)$ is closed $\implies$ $A$ is bounded. Clearly $X(A)$ is closed $\implies$ $A$ is closed because $A$ is an intersection of $X(A)$ with a ...


1

Another example is $$\{0\}\cup\biggl\{\frac1n: n\in\mathbb N\biggr\}$$ Maybe that was expected?


1

As stated, your answer appears correct, but I wonder what I would think if I saw all of what you wrote. You say each neighborhood of $3$ contains only $3$, and that's wrong: There is in fact only one open neighborhood of $3$ that contains only that point, and that neighborhood is $\{3\}$. The set $(1/2 - 1/100,1/2)\cup\{3\}$ is another open neighborhood of ...



Only top voted, non community-wiki answers of a minimum length are eligible