# Tag Info

5

It is the preimage of the closed set $\{1\}$ under the continuous map $A\mapsto \det A$. For $n=1$ you may try to enumerate all elements

5

First let me tell you the flaws in your argument. 1) nobody said anything about $X$ being connected. 2) You never used compactness of $X$. 3) You are using a property of "real" valued functions, these are metric spaces, which are much more general than real spaces. Here's how you do it. Take any open covering for $f(F)$ in $Y$, say $\{U_\alpha\}_{\alpha \in ... 4 You might want to proceed in the following manner: Closed subsets of compact sets are compact, thus$F$is compact; Continuous image of a compact set is compact, hence$f(F)$is compact in$Y$; Compact sets in a metric space are closed, hence$f(F)$is closed. 4 Don't have any ideas on your proposed proof, other than that already suggested by Mr. Sinclair, namely that whatever argument you use will certainly invoke compactness. What I can do is add to your collection of proofs. You know a Bolzano-Weierstrass argument and Heine-Borel argument for Dini's theorem. In Exercise 9.4.7 Elementary Real Analysis I ... 4 The equivalence between compactness and "closed and totally bounded" is only true in complete metric spaces. Since$\Bbb Q$is not complete, you need to require more than closure, you need to require completeness. Namely, a metric space$(X,d)$is compact if and only if it is complete and totally bounded. You are absolutely right that$\Bbb Q\cap[0,1]$is ... 4 Colimits in the category of open subsets (and inclusions) are just unions. A filtered subcategory of the category of open subsets has the property that any finite union of objects is contained in another object. So if$A$is a compact open set then if there's a morphism to the union of the open subsets in a filtered subcategory (i.e., if$A$is a subset of ... 3 Another reason: a compact subset of the reals is closed and bounded (Heine-Borel theorem), and the rationals are distinctly not bounded. 3 Quoting from the paper (slightly modifying notation): "now let$\{F_i: i \in I\}$be a collection of closed sets with the FIP. Each$F_i$consists of all ultrafilters that include all of a collection$\mathcal{F}_i$of subsets of$\mathbb{N}$". (this is hopefully clear, this is argued in the previous paragraph). So we know$F_i = \{ \mathcal{F} \in ...

3

It is available on Gallica (the online platform of the France National Library) here.

3

Note that $+\colon \Bbb R^2\to\Bbb R$ is continuous, hence the image $C$ of the compact set $A\times B\subset \Bbb R^2$ is compact.

3

You're quite right - every nonstandard model of $Th(\mathbb{N})$ has lots of infinite elements! For example, if $\mathcal{N}$ is a nonstandard model and $a\in\mathcal{N}$ is an infinite element, then: $a^2, a^3,...$ all exist in $\mathcal{N}$ and are infinite. A number $b$ such that either $2b=a$ or $2b+1=a$ exists (depending on the parity of $a$ in ...

3

Just show that $\{x_n:n\in\mathbb{N}\}$ is discrete, i.e. every singleton is an open set. For each $n\in\mathbb{N}$, show that $B(x_n,\epsilon)$ only contains $x_n$.

3

Suppose not. Then there is an infinite sequence of $F_n$ such that all the $F_n$ aren't contained inside $G$. Since $G$ is open, each $F_n \setminus G$ is also closed; each is also nonempty by assumption. Consider the intersection of those.

3

Induction proves the statement for any finite number, not a countably infinite number. The reason it only shows it for finite numbers is that induction proves that $P(k)$ is true, then so is $P(k+1)$. Also note that if you find a counterexample to a statement, then the statement can't be true (in particular, you found a counterexample to your statement, so ...

3

Let $(a_n,b_n) = (x_n - 2^{-n - 2} \epsilon, x_n + 2^{-n - 2} \epsilon)$. Then $x_n \in (a_n,b_n)$ and $\sum b_n - a_n < \epsilon$.

3

Every open cover of the union is an open cover of each set. So for each set there is a finite subcover. The union of the finite subcover is still finite and covers the union of the two sets. So the union is indeed compact.

3

The above is correct. To see that it is indeed a cover, let $x \in [1,3)$. As $x < 3$, we know that $3 - x > 0$ and so for large enough $N$ we know that $\frac{1}{N} < 3 -x$, or $x < 3 - \frac{1}{N}$, so that $x \in \left[1,3-\frac{1}{N}\right)$.

2

Because every sequence is a function from $\mathbb N \to \mathbb R$ so in other words you want to find the measure of $\mathbb N$ in $\mathbb R$ and every countable set has measure $0$ i.e $\mathbb N$ can be easily covered by open balls of radius $\epsilon$ by $\mathbb N=\cup _{n\in \mathbb n}(n-\epsilon,n+\epsilon)$ for each $\epsilon>0$

2

A slightly different answer using sequences. Take $F$ a closed set of $X$. Now take a sequence $(y_n)$ of elements in $f(F)$ converging to $y\in Y$. We want to show that $y\in f(F)$. Because $y_n\in f(F)$ we can find $x_n\in F$ such that $f(x_n)=y_n$. Now $(x_n)$ is a sequence of $X$ (compact), so it admits a subsequence converging $(x_{\varphi(n)})$ to ...

2

There is only one compatible uniformity with the topology on $L$. This follows from this paper or the well-known result (reference anyone?) that locally compact Tychonoff spaces have a unique uniformity iff the one-point compactification is the same as the Cech-Stone compactification. So $L$ has a unique uniformity that is totally bounded and not complete. ...

2

First I'll consider your (bolded) question about taking only discs with centers in $S^1$. Indeed we can do this - consider your open cover $\{ U_i\}$. Each $x\in S^1$ is contained in some $U_x\in\{ U_i\}$, and $U_x$ is open, so it contains some open ball in $\mathbb{R}^2$ centered at $x$, say $B_x$. Indeed, it should be clear that the collection of open ...

2

Nope! Because there are 2 items in your cover and that is a countable sub cover. As every singleton set is closed we could consider our cover to be the union of all singleton set for real numbers in $[0,1]$. As the interval $[0,1]$ is not countable over the reals and removing any one of the singleton sets makes it no longer a cover this satisfies the ...

2

HINT: Let $p$ be the point at the tip of the cone. Show that every nbhd of $p$ contains a set of the form $$q\left[\bigcup_{n\in\Bbb Z}\big(\{n\}\times[a_n,1]\big)\right]\;,\tag{1}$$ where $q$ is the quotient map, and $a_n\in[0,1)$ for each $n\in\Bbb Z$. Such sets are closed, so if the cone were locally compact, one of them would have to be compact. But ...

2

[Continued: ... indulge me.] There is agreement now that this question is really about Dini's theorem and compactness arguments. Mr. Khor wanted to try for a proof based on using a single modulus of continuity for the whole sequence, not to avoid compactness arguments--just for a different perspective. I think all proofs of Dini's theorem look the same ...

2

This operator is known as the Volterra operator. As Martin R pointed out, its kernel is trivial. The range can't be described in simpler terms than "antiderivatives of $L^2$ functions", which is a tautology. The operator is compact. One way to show it is to apply a general theorem saying that all Hilbert-Schmidt operators are compact (as is gone here). ...

2

So you have a set of the form $$C=\{x\in\ell_2; |x_n|\le a_n\}$$ where $a_n$ is a series of positive real numbers such that $\sum a_n^2<+\infty$. You want to show that this space is totally bounded. Let $\varepsilon>0$. We want to show that there exist finitely many points in $C$ such that each point of $C$ is within the distance $\varepsilon$ from ...

2

Assume that the period equals $p$ and use the fact that the function attains all its values on, e.g., $[0,p]$, which is compact.

2

Since $f$ has a finite limit at each point of $K$, for each point $x_0\in K$, there is a $\delta>0$ such that $|f(x)-L|<1$, i.e. $f$ is bounded in $(x_0-\delta, x_0+\delta)$. Since it holds for any point $x_0\in K$, $K\subset\bigcup_{x_0\in K}(x_0-\delta, x_0+\delta)$, i.e. an open cover of $K$. Since $K$ is compact, there is a finite subcover that ...

2

A compact space is complete. Another reason: a compact subspace is closed. And precisely, the closure of $\mathbf Q$ is $\mathbf R$.

2

Your set is $f^{-1}[0,1]$. As $f$ is continuous, inverse image of closed set is closed. So $f^{-1}[0,1]$ is a closed subset of $D$ and as closed subset of compact Hausdorff space is also compact, you will have $f^{-1}[0,1]$ is compact. alternative: Suffices to show every infinite set in $f^{-1}[0,1]$ has a limit point in $f^{-1}[0,1]$. Let $x_n$ be a ...

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