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0

It seems the following. We don’t have much choice to find this subgroup. Let $Z$ be a subgroup of the group $f(H)$, topologically isomorphic to $\Bbb Z$. Put $D=\ker f$ and $C=f^{-1}(Z)$. Since the map $f$ is continuous, the group $C$ is a closed subgroup of a locally compact group $G$. So $C$ is a locally compact group too. By Theorem 2.3 from [HC], every ...


1

Consider the $f(x) = \lceil|x|\rceil$, that is rounding up the absolute value, which is surjective. Then, $f$ sends bounded (compact or not) subsets of $\mathbb R$ to finite (and thus the only compact) subsets of $\mathbb N$. An Example for non compact image: Let $f(x)$ be the number of decimal digits of $x$ if $x$ has an finite decimal expansion, ...


1

You could define $f:\mathbb R\rightarrow\mathbb Z$ such that $f(x)=\lfloor x\rfloor$. Since every compact subset of the reals has only a finite number of integers in it, this sends compact sets to compact sets. You could define $f:\mathbb R\rightarrow\mathbb Q$ be letting $f(x)$ be the closest member of the (compact) sequence $\{1/n\}\cup\{0\}$ (choose the ...


1

The region $xy>0$ is just the union of (1) the interior of the first quadrant, and (2) the interior of the third quadrant. That is the set you're trying to prove to be open. If $(x,y)$ is in the first quadrant, it distance from the $x$-axis is $y$ and its distance from the $y$-axis is $x$. Make the radius of the disk less than or equal to both of those ...


0

It might also help to think about this visually. On the cartesian plane, your set represents quadrants $I$ and $III$, not including the $y$-axis or the $x$-axis. I recommend drawing this on paper, and then tossing a few points in quadrants $I$ and $III$. Next you'll want to figure out a way to guarantee that you can enclose the points in an open ball. One ...


0

As you mentioned correctly one has to show, that for every $(x,y)\in M$ (I called the set M) one has to find a radius $r$ such that $B_r \subset M$. So let $(x,y)\in M$ be arbitratry chosen, then if you choose $r=\min(|x|,|y|)/2$. $xy>0$ implies that $x$ and $y$ have the same signs. So wlog $x>0$ and $y>0$. Because of the choice of $r$ is $y-r>0$ ...


2

If you can use continuous functions, then the set in question is the inverse image of the open interval $(0,\infty)$ under the continuous function $\mathbb R^2 \to \mathbb R$ with $(x,y)\mapsto xy$.


1

Draw the set in question. It has an easy-to-describe boundary, $B$; given any point $(x, y)$ in the set, it should visually be easy to find the distance from $(x, y)$ to $B$. Any radius smaller than this distance will work.


2

Prove it using a different approach: Let this set be named $A$. Then: $$A^C = \{(x,y) \in \mathbb R^2, \ xy \le 0 \}$$ Let $\{(x_n,y_n)\}_n$ is a sequence in $A^C$ converging to $(x,y)$. We have $x_n y_n \le 0$, $\forall n$ $\implies \lim(x_n y_n) \le 0$, i.e. $xy \le 0$. Then, $(x,y) \in A^C$. This shows that $A^C$ is closed, that's, $A$ is open.


1

Hint: Consider the maps \begin{alignat*}{2} G&\longrightarrow G\times G&&\longrightarrow G\\ x&\longmapsto(x,1)&&\longmapsto x^{-1}\cdot 1 \end{alignat*} then \begin{alignat*}{2} G\times G&\longrightarrow G\times G&&\longrightarrow G\\ (x,y)&\longmapsto(x^{-1},y)&&\longmapsto (x^{-1})^{-1}y=xy\\ \end{alignat*} ...


3

We have $$f(X)\subseteq\{0\}\cup f(\mathrm{supp}(f)),$$ which is compact in $\mathbb{R}$ (since $\mathrm{supp}(f)$ is compact and $f$ is continuous), hence bounded.


5

Hint: (1) First, show that, or note that $x\mapsto (x,e)$ is continuous. (2) Show that $x\mapsto x^{-1}$ is continuous, by showing that it is the composition of two continuous map. (3) Show that $(x,y)\mapsto xy$ is continuous, by the same method of (2).


1

If you do know the Weierstraß theorem, then you can prove it like that: Let $f \in \mathcal{K}(X)$ and denote by $K$ the support of $f$. Then $f|_{K^c}=0$ by the very definition of the support. Moreover, by the Weierstraß theorem, $f|_K$ is bounded. Combining both facts, proves that $f$ is bounded. If you do not know the Weierstraß theorem, then have e.g. ...


1

Here is the case for $d=2$. It extends easily to higher dimensions. I will assume that $K$ is compact for if $K$ is bounded then it is contained in a compact ball $\subset \mathbb R^{2}$. Note that $f'$ is continuous there, by assumption. Now, the derivative of $f$ is the linear transformation $f':\mathbb R^{2}\rightarrow \mathbb R$: $h=(h_{1},h_{2}) ...


1

It's not true -- for a counterexample, let $d=1$, $K=[0,\infty)$, $f(x)=x^2$. The major problem with your proof is that you don't have any assumptions that will guarantee that $K$ is contained in any compact subset of $\mathbb R^d$. You also seem to assume that $\|x-y\|\ge 1$, which will not be true in general. So even if you have a $K'$, what you need to ...


0

Let us define the shift in x operator: \begin{eqnarray} S_x : \mathbb{R}^2 &\to& \mathbb{R}^2 \\ (x,y) &\mapsto& (x + x, y+x) \end{eqnarray} It is easy to prove that this is continuous using open balls. Pick $(x,y) \in \mathbb{R}$ then $S_x(x)=(2x, x+y)$. An arbitrary open ball around $S_x[(x,y)]$ is $B[ (2x, x+y), r]$, Now ...


2

In the quoted passage no assertion is made that such a set actually exists. Read "let x be Y" as "assume for the sake of argument that some object called x satisfying Y exists". In general, this can go several places. It can lead to a contradiction, proving that such an x fails to exist. Or it could show some property that all such objects must have, even ...


1

A manifold is a space $X$ such that each point $p\in X$ has neighborhoods which are homeomorphic to open euclidean balls. In particular the neighborhoods of all points $p\in X$ "look alike". If the manifold $X$ is a compact topological space then $X$ is called a closed manifold (this wording is somewhat unfortunate, see below). The prime examples of ...


0

Note well that the answer is necessarily positive only in finite dimensions, I.e. Balls in R^n. In infinite dimensions, it depends on the topology used. For example, in infinite dimensional Hilbert space, the unit ball is non-compact in the topology induced by the norm. This can be seen as a basis can be viewed as a sequence without accumulation point. If ...


1

The argument does not go through. A quite complicated construction called the “irrational-slope topology” is given by Steen and Seebach (1978) with the following properties: Hausdorff; but not regular; yet second-countable (and hence Lindelöf). I think the problem is as follows. In proving that every Lindelöf regular space is normal, you first take two ...


3

I think I can confirm your suspicion that this doesn't necessarily hold if the target space is non-Hausdorff, assuming I haven't made a mistake somewhere... Let $\mathbb{R}$ be the real line in its standard topology. Let $\mathbb{R}_0$ be the real line with the topology whose non-empty open sets $U$ are precisely the standard open sets $U \subseteq ...


0

Since this problem is a bit trivial (one condition defining the set implies the other), let us consider a more general scenario where the second condition is $ax+by < c$ for some $a,b,c \in \mathbb{R}.$ It is easy to see that the set $X := \{ x^2 + y^2 < 1 \}$ is open using the open ball definition, since it's already an open ball of radius $1.$ To ...


0

The function $d(x,y)$, defined on $K \times K \rightarrow \mathbb R$ , is a continuous (uniformly-continuous, actually) Real-valued function, defined on the compact set $K\times K$. A continuous function on a compact set reaches , takes on a (finite) maximum value, say $M$. This value can serve as the diameter of a ball containing $K$. And re the issue of ...


1

You need to prove the converse now, i.e, if $\prod_{\alpha} U_{\alpha}$ is open, then $U_{\alpha} = X_{\alpha}$ for all but finitely many $\alpha \in A$. We will argue by contradiction, so suppose $U_{\alpha} \neq X_{\alpha}$ for infinitely many $\alpha$, and also that $\prod_{\alpha} U_{\alpha}=:U$ is open. Since any open set is a union of basis sets, it ...


1

The intersection of two open sets is open. Or, straightforwardly, since $\{x^2+y^2<1|(x,y)\in\mathbb{R}^2\}$ is contained in $\{x+y<3\}$, the set $$ \{x^2+y^2<1, x+y<3|(x,y)\in\mathbb{R}^2\}=\{x^2+y^2<1|(x,y)\in\mathbb{R}^2\} $$ is open.


2

The set is just $$\{x^{2}+y^{2}<1\}$$ So it's open.


1

The usual strategy is to let the open cover of the compact set $K$ be $\mathcal B=\{B(x,1):x\in K\}$. Since each point in $K$ has its own ball, $\mathcal B$ definitely covers $K$. Since $K$ is compact, there exists a finite subcover.


0

HINT: With a little more work you can choose the sets $U_n\in\tau$ so that $U_{n+1}\subseteq\operatorname{cl}_\tau U_{n+1}\subseteq U_n$ for each $n\in\Bbb Z^+$. Now show that $\bigcap_{n\in\Bbb Z^+}U_n$ is clopen in $\tau_\delta$, and use it to get a $\{0,1\}$-valued continuous function separating $x$ from $F$ in $\langle X,\tau_\delta\rangle$. (In case ...


2

From the introduction: As will be seen from the list of contents the articles cover a wide range of topics. Some are more technical than others, but the reader without a great deal of technical knowledge should still find most of the articles accessible.


2

HINT: For the first question suppose that $U\subseteq X$ is such that $U$ is both open and closed. Use the fact that $\pi^{-1}[\{y\}]$ is connected for each $y\in Y$ to show that for each $y\in Y$, either $\pi^{-1}[\{y\}]\subseteq U$, or $\pi^{-1}[\{y\}]\subseteq X\setminus U$. Then use the fact that $Y$ is connected and $\pi$ is a quotient map to show ...


2

We can use compactness of the product $K\times K$ without mentioning it. There is a sequence $(x_m,y_m)\in K\times K$ such that $$ \lim_{m\to\infty}|x_m-y_m|=\sup_{x,y\in K}|x-y| $$ by the properties of the supremum. Since $K$ is compact, the sequence $(x_m)$ has a convergent subsequence, say $(x_{m_h})$. Similarly, the sequence $(y_{m_h})$ has a convergent ...


2

Let $y=x^{2}$. Consider $f(x,x^{2})=\frac{x^{4}}{2x^{4}}=\frac{1}{2}$.So it's not continuous at $(0,0)$. (Even it does not have a limit, you can plug $y=0$)


1

Let $A$ be a closed set in $X$. Let $a \in A$. Now, since $X$ is locally compact, we can find a compact neighborhood $V$ of $a$ in $X$. $V \cap A$ is closed in $V$, and closed subsets of compact sets are compact.


1

A two variable function can be thought of as a one-variable function on a product space. For example, if $X,Y,Z$ are spaces, and $f(x,y)$ is a two-variable function taking values in $Z$, for $x\in X,y\in Y$, then we think of $f$ as a function $f\colon X\times Y\to Z$. This leaves the matter of what metric/topology to put on the product $X\times Y$. We ...


2

I'll answer your second question first. You have $f:K\times K\to\mathbb R$. The space $K\times K$ is the product of two compact sets, and is therefore compact. This is not hard to prove, and is a very simple result of the Tychonoff Theorem. Now, in order to apply the extreme value theorem, we have to show that $f$ is continuous. To do this, pick ...


1

Let $S$ be an uncountable subset of $T$, and let $D$ be a countable subset of $S$; it suffices to prove that $D$ is not dense in $S$. Since $D$ is countable, there is an $\alpha<\omega_1$ such that $D\subseteq\bigcup_{\xi<\alpha}\operatorname{Lev}_\xi(T)$. Now use a cardinality argument to show that there are $x,y\in S$ such that ...


2

With a little extra work an Aronszajn line does what’s wanted. If $\langle X,\le\rangle$ is an Aronszajn line, define a relation $\sim$ on $X$ by setting $x\sim y$ iff the interval between $x$ and $y$ is countable. It’s easy to check that $\sim$ is an equivalence relation with order-convex equivalence classes, so that $X/\!\!\sim$ naturally inherits a linear ...


0

Let U ⊂ R, and let f: U ⊂ R $\to R^2$ be t $\to$ ($t^2$ - 1 , t($t^2$ - 1)). It is locally injective at all points of R, but not injective in any interval (-a,a) with $a\gt 1$, because f(-1)=f(1)=(0,0)


3

Yes. Say $E_0\in\mathcal O$. Let $\mathcal O_0$ denote the set of $E\in\mathcal O$ such that $E$ is connected to $E_0$ by a chain of vertices in that graph. Let $\mathcal O_1=\mathcal O\setminus \mathcal O_0$. Let $V_j$ be the union of the $E\in \mathcal O_j$, $j=0,1$. So the $V_j$ are open and $X=V_0\cup V_1$. Now if $E\in\mathcal O_0$, $F\in\mathcal O$ ...


3

This seems to be true. Note that if $K \subset \mathcal{O}$ then $\bigcup K$ is an open subset of $X$. Also, if $K_1$ and $K_2$ are (the vertex sets of) distinct components of $\mathcal{O}$, then $\bigcup K_1 \cap \bigcup K_2 = \emptyset$. Thus the decomposition of $\mathcal{O}$ into components induces a decomposition of $X$ into open sets.


0

As already noted in a comment by "1999", this is not generally true as written, even for Euclidean spaces. For example take $X = \mathbb{R}$ and $U = \{ (t, x) \in [0,1] \times X \mid t < |x| \}$. However, with some minor modifications we get a true statement. Let $X$ be a perfectly normal space and $U$ an open subset of $[0, 1] \times X$. Then ...


7

Hint. Consider the set $S=\{k+2\pi u | (k,u) \in \mathbb{Z}^2\}$. S is an additive subgroup of the reals. And you probably know that the additive subgroups of the reals are either closed (and have a least positive element) or dense. $S$ cannot be closed as this would imply that $\pi$ would be rational. So $S$ is dense. As $\sin$ is continuous, $\sin(S)$ is ...


2

Related to some of the comments, you might be interested to know that $(-\infty,\infty]$ in the cyclic order topology is homeomorphic to $S^1$. The cyclic order topology is coarser than the corresponding linear order topology, in this case strictly coarser.


4

If $(-\infty,+\infty]$ is construed as a mere set rather than as a topological space, then it's not homeomorphic to anything including itself. Now lets add a topology: a basic open neighborhood of a point other than $+\infty$ is an open interval containing that point; a basic open neighborhood of $+\infty$ is a set of the form $(a,+\infty]\cup (-\infty,b)$; ...


9

Neither is homeomorphic to $S^1$. $(-\infty,\infty]$ is not compact, while $S^1$ is, so they cannot be homeomorphic. $[-\infty,\infty]$ is compact, being homeomorphic to $[0,1]$, but removing any one point of $\Bbb R$ from $[-\infty,\infty]$ disconnects it, while removing one point from $S^1$ does not disconnect $S^1$, so they are not homeomorphic either.


2

A lot of times these two mean the same thing, but it is important to consider the superset of which this is an interval. Sometimes, (especially in measure theory, which is why I mention it) it is useful to work in the extended reals, which includes a point at $\infty$, so $(a,\infty)$ means every number greater than $a$ accept infinity and $(a,\infty]$ ...


4

\begin{align*} (a,\infty)=&\,\{x\in\mathbb R\,|\,x>a\},\\ (a,\infty]=&\,\{x\in\mathbb R\,|\,x>a\}\cup\{\infty\}. \end{align*} The latter set includes an extra point termed “positive infinity.” Note that it is not a real number, but in certain areas of mathematics, especially in measure theory, it is useful to extend $\mathbb R$ by this single ...


2

There are many equivalent ways to construct the topology on $C^\infty_c(\Bbb{R^d})$ that makes it an LF-space. I'm not familiar with Tao's method, but I've looked at a different construction that should be equivalent, and is in my opinion very intuitive. First, we set our goal to be to find a topology on $C^\infty_c(\Bbb{R^d})$ that makes it a locally ...


2

No compact subset of $\mathbb R^n$ is open except $\varnothing$. Since $\mathbb R^n$ is a connected space, its only subsets that are both open and closed are $\varnothing$ and $\mathbb R^n$, and the latter is not compact. Lots of sets are both open and bounded. Every open ball is both open and bounded.


5

Note well that the only subsets of $\mathbb R^n$ ($n\in\mathbb N$) that are both closed and open are the whole set and the empty set. (This is the consequence of a fundamental property of Euclidean spaces called connectedness, which is the topological formalization of the intuitive geometric notion of contiguity.) Since $\mathbb R^n$ is not compact (it's ...



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