New answers tagged

1

I would suggest a different way of attacking this one: Define the following, for $\epsilon>0$: $B_\epsilon(x, \mathbb{R})= \left\{ y \in \mathbb{R}: |x-y|<\epsilon \right\}$ $B_\epsilon(x, \mathbb{Q})= \left\{y \in \mathbb{Q}: |x-y|< \epsilon \right\}$ It is then clear that $X \subseteq \mathbb{R}$ is open if and only if for each $x \in X$ there ...


1

It’s not true that all open sets are of the form $X\setminus F$, where $F$ is finite; what is true is that every set of that form is open. HINT: For each $x\in X$ let $U_x=X\setminus\{x\}$. Because $X$ is $T_1$, you know that each $U_x$ is open. Use the fact that $X$ has no isolated points to show that each $U_x$ is also dense. Then consider $\bigcap_{x\in ...


2

Perhaps the most elementary proof is the one that I first encountered as a freshman, using the Alexander subbase lemma. It requires Zorn’s lemma, but it does not require knowledge of filters, ultrafilters or nets. It’s carried out completely in this PDF. (And Alexander’s result is of some interest in its own right.)


0

NO. Let $V=[0,1)$. Let $p=<-1,0>\in R_l^2.$ We have $f(p)=0 . $ Suppose $U$ is a nbhd of $p$ such that $$\forall q\in U\; ( f(q)\in V). $$ We have $U\supset [-1,-1+r)\times [0,s)$ for some $r,s \in R^+ . $ But then $q=<-1,s/2>\in U$ and $f(q)=-s/2\not \in V, $ a contradiction. Good Q nevertheless.


2

In fact $$(\mathbb R\times\mathbb Q)\setminus(\mathbb Q\times\mathbb Q)=(\mathbb R\setminus\mathbb Q)\times\mathbb Q,$$ the reason being that if $x\notin\mathbb Q,$ then $(x,y)\notin(\mathbb Q\times\mathbb Q)$ whatever $y$ is. Hope this helps. Edit: For $(x,y)\in(\mathbb R\times\mathbb Q),$ by definition, $(x,y)\in(\mathbb Q\times\mathbb Q)\iff ...


2

Spoted a mistake. It should be $$(\mathbb{R} \times\mathbb{Q} ) \backslash (\mathbb{Q} \times \mathbb{Q})=(\mathbb{R} \backslash \mathbb{Q}) \times \mathbb{Q}$$


2

No. A separable metric space $X$ is hereditarily Lindelöf (even second countable). And this implies that every uncountable subset $A$ of $X$ has a point $p$ such that every ball around $p$ intersects $A$ in uncountably many points (otherwise every point is locally countable and we use the Lindelöfness of $A$ to get a contradiction). And this allows us to get ...


4

Take any periodic point $q_1$. Its orbit is a finite set, and therefore is not dense. So there must be another periodic point $q_2$ whose orbit is not $\mathcal O(q_1, f)$. But the orbits of periodic points are either the same or disjoint.


1

If $f$ is open then for every $n$, the set $B_{\frac{1}{n}}(0)$ is open in $N$ and so $f[B_{\frac{1}{n}}(0)]$ is open in $N_1$ and it contains $0 = f(0)$. So $0$ is an interior point of $f[B_{\frac{1}{n}}(0)]$ which means that we have some $r>0$ such that $B_r(0) \subseteq f[B_{\frac{1}{n}}(0)]$. This shows necessity of the condition. Suppose $f$ obeys ...


1

You have to show that the image of the open subset $U$ is open, let $x\in U$, there exists $n$ such that $B(x,1/n)\subset U$. Consider $r$ such that $B(0,r)\subset f(B(0,1/n)$, $B(f(x),r)=f(x)+B(0,r)\subset f(x)+f(B(0,1/n))=f(B(x,1/n))\subset f(U)$. This implies that $f$ is open.


1

This is functional in general. Let $X$ be a set an let $\mathcal V\subseteq\wp(X)$. Then the collection of finite intersections of sets in $\mathcal V$ automatically has the basic properties of a base of a topology. This in the understanding that the empty intersection equals $X$. Denoting the collection by $\mathcal{V}^{\stackrel{\cap}{f}}$ we have ...


1

We have: $$\left(\bigcup_{i \in I} A_i\right) \cap \left(\bigcup_{j \in J} B_j\right) = \bigcup_{(i,j) \in I \times J} (A_i \cap B_j)$$ so that the intersection of two unioned families is again a unioned family, over a bigger index set, and if the $A_i,B_j$ come from a family which is closed under finite intersections, the latter is also a union from that. ...


2

No, having a CW complex structure is absolutely not enough to be a manifold. Being a CW complex is very easy, being a manifold is hard. Take the wedge sum of two circles $S^1 \vee S^1$ for example: it's a CW complex, but not a manifold. If a space $M$ is a compact manifold without boundary, then you can read its dimension using singular homology: it is the ...


2

It is true that $(0,1)$ is totally bounded, and this is a property of the metric (or of the uniformity, if you know about those), not of the topology. So it is not necessarily preserved by homeomorphisms (but it is by isometries!), as your own example in fact shows. That $(0,1)$ is totally bounded can be seen in different ways: for $r>0$ we can cover ...


1

The set $X$ is the unit sphere in $\Bbb{R}^4$. Any two distinct $x,y\in\Bbb{R}^4$ are contained in some plane $Y\subset\Bbb{R}^4$, and hence $x,y\in X\cap Y$. But $X\cap Y$ is then a circle in $Y$, which is of course path connected. So $x$ and $y$ are connected by a path in $X\cap Y\subset X$, hence $X$ is path connected.


2

This follows from the standard result given in, for example, Topology and Groupoids: $4.3.2$ Let $f : X \to Y$ be an identification map and let $B$ be locally compact. Then $$f × 1 : X × B \to Y × B$$ is an identification map. The definition of locally compact here is that each point has a base of compact neighbourhoods. This inconvenient ...


-2

the assumption that since its finite thus countable is false. Countable means a bijection with N. It can be infinite (countably infinite, or simply just stated countable)


0

For any Banach space $X$, if there is an exhaustion $X=\bigcup_n K_n$ by compact sets, then one of the $K_n$ has an interior point by Baire's theorem. This implies that $X$ is finite dimensional.


1

You're right about the final point: in the (non-compact!) space $(0,1)^{[0,1]}$ (in the product topology) the subset $[\frac{1}{10},\frac{1}{2}]^{[0,1]}$ is indeed compact (by the same Tychonoff theorem); it is not true that we can only have a finite product of "compact component spaces"; you seem to be confused with open sets, where basic open sets depend ...


1

For $a \implies b$, consider a subset $X \subseteq \Bbb R^n$, and the collection of functions $$ f_1(\mathbf x) = x_1, f_2(\mathbf x) = x_2,\dots, f_n(\mathbf x) = x_{n} $$ Note that $\mathcal A$ contains the constant function $$ g(x) = (f_1)^0(f_2)^0 \cdots (f_n)^0 $$ moreover, $\mathcal A$ separates points since for any two elements $\mathbf a = ...


0

$f(x)$ is a function which takes $x$ to a function which maps $t \mapsto t(x(t)+1)$. This is sometimes called "currying".


0

It is usually better to start simplifying the hypothesis as much as you can. For example, start by taking a finite subcover of the $U_\alpha$. Also, think about the case where $U_\alpha=X$. Even though the result is trivial in this case, the procedure given by a general proof should work in this particular case. When we take functions $h_i$ but simply have ...


1

I think the answer is no. Each $K_i$ is separable, hence $\cup K_i$ is separable. But $C_b(\mathbb R^n)$ is nonseparable. To see the latter, think of a countable family of closed pairwise disjoint balls $B_m$ and continuous functions $f_m$ such that for every $m$: i)the support of $f_m$ is contained in $B_m;$ ii) $0\le f_m\le 1;$ iii) $f_m = 1$ somewhere in ...


2

I think this is related (or maybe follows from) the result by E. Michael that the product of a locally compact space and a $k$-space is again a $k$-space.


4

Necessary and sufficient conditions are that $A\supseteq\overline{A^\circ}$, $\overline{A^\circ}$ is compact, and $A^\circ$ is homeomorphic to $\mathbb{R}^n$. First, suppose these conditions hold. Note that $A/{\sim}\cong\overline{A^\circ}/{\sim}$, so we may assume $A=\overline{A^\circ}$ and in particular that $A$ is compact. Then since $\sim$ is a closed ...


6

Your picture of the surface of a donut is correct: it is curved. Not all metrics on a topological surface will have the same local properties. For example, although a donut surface does indeed have flat metrics, it also has nonflat metrics, and the donut surface is one of them. 2d entities living on the surface of a donut could indeed distinguish their ...


1

The proof is essentially OK. Minor quibbles: you don't need the $\{\emptyset\}$, in the definition of $\mathcal{A}_\mathcal{B}$, because we get it for free using $\mathcal{C} = \emptyset$ which has as its union $\emptyset$ as well. I suppose you already know that $\mathcal{A}_\mathcal{B}$ is a topology, either by the definition of a base or a theorem, and ...


1

If you wish to construct product measure with infinitely many probability spaces as factors, then no topological assumptions are needed. More precisely, if $\{(E_t,\mathcal E_t,\Bbb P_t): t\in T\}$ is a non-empty collection of probability spaces indexed by some set $T$, then there is a unique probability measure $\Bbb P$ on the product space $(\times_{t\in ...


2

Consider $A=(0,1)\cup (1,2)$ which is open w.r.t. the usual topology on $\mathbb{R}$. Then, $\bar{A}=[0,2]$, so int$\bar{A}=(0,2)\not= A$


3

Yes, the operations you list are all open maps. Proving that the image of each element of a basis of the domain maps to an open set is sufficient, since every open set is a union of the basis elements, so the image of any open set is a union of the images of basis elements, each of which was open. I don't think there's any better or more general approach ...


4

Are you familiar with this picture? (courtesy of Wikimedia) To get a torus you start with a full square $[0,1]^2$ and then you identify opposite edges in the way depicted above. Indeed if you first identify the red edges you get a cylinder, and then the blue edges become the top and the bottom circle. If $x$ is for blue and $y$ is for red, the ...


2

For the first two use members of $\Bbb Q^{\Bbb N}$ and $(\Bbb Q\cap[0,1])^{\Bbb N}$ that are $0$ for almost all $n$. For the last just use sequences that are $0$ for almost all $n$.


1

You can actually prove something a little bit more general: Let $X$ be a topological space and $A_1$ and $A_2$ be subspaces of $X$. Let $g_{i}:A_i\rightarrow Y$ be continous for $i=1,2$. If $g_1$ and $g_2$ coincide on $A_1\cap A_2$ and $A_1$ and $A_2$ are either both open or both closed, then the function $f:X\rightarrow Y$ defined by $f(x)=g_1(x)$ if $x ...


3

For the case that $A$ and $B$ are open : If $U \subset Y$ is open, then $f^{-1}(U) = (f^{-1}(U) \cap A) \cup (f^{-1}(U) \cap B)$ which is open since the restrictions are continuous.


0

(one year later) I think I have a valid proof now: (1). The def'n of the topology $T_d$ generated by a metric $d$ on a set $S$ is the topology whose base is the set of open $d$-balls. (2). Two metrics $d,d'$ on $S$ are inequivalent (meaning that $T_d\ne T_{d'}$) iff there exists $U\subset S$ which belongs to one of $T_d,T_{d'}$ but not the other. ...


1

The problem is not to construct an infinite product of sigma-algebras. This always exists and is defined as you have stated, as the smallest sigma-algebra that makes all the projections measurable. The problem is to find measures on this space. Of course a measure on the infinite product always pushes forward to a measure on the finite products contained in ...


0

The following theorem is given with proof here: Let $X$ be a topological space that is path-connected and locally path-connected. Let $G < \text{Hom}(X)$ be a group of homeomorphisms on $X$. Then the projection map $q: X \to X/G$ is a covering map if and only if $G$ acts properly discontinuously on $X$. Properly discontinuously means that for every $x ...


1

I think the key here is showing $f(x) = (f_1(x),f_2(x))$ is continuous if $f_1$ and $f_2$ are. You'll want to invoke the characteristic property for both canonical projections: (Note that this is analogous to Lee's proof for the product map $f_1 \times f_2$, but our $f$ is different from this product map.) Combine this with the continuity of $x+y$ and ...


2

Let $(X, d)$ be a metric space, $S \subset X$, and $f:S \longrightarrow S$ a function be such that $d(f(x), f(y)) \leq c d(x, y)$, for all $x, y \in S$, where $0 \leq c < 1$ is given. Fix $\epsilon > 0$ and choose $a \in S$. The case where $c = 0$ is trivial. Assume $c > 0$ and let $\delta = \epsilon / c$. For all $x \in S$ with $d(x, a) < ...


3

The Möbius band has a boundary, and therefore doesn't give a contradiction to the statement.


0

It is true that finite sets are closed in every T$_1$ space, and thus they are closed in every discrete space. Also by the definition of the discrete topology, $\textit{every}$ subset of the space is open. So suppose $X$ is discrete and has more than one point. Let $x\in X$. Then $\{x\}$ is open. It is also closed (it is finite), and so its complement is ...


3

Let $B$ be given. Let $U=P\setminus \overline B$. Let $\{p_i:i\in\omega\}$ be a dense subset of $U$. Let $D_0$ be an open disk with $p_0\in D_0\subseteq U$. Let $i_1=\min\{i\in\omega:p_i\notin \overline D_0\}$. Let $D_1$ be an open disk with $p_{i_1}\in D_1\subseteq U$ and $D_1\cap D_0=\varnothing$. Continue this process, recursively defining a collection of ...


1

Open sets: Intuitively -- open sets have a soft edge. Mathematically -- for every point in an open set there exists a neighborhood about that point where every point in that neigborhood is in the set. Closed sets: Intuition -- Have hard edges. Math -- Are the complements to open sets. Closed sets contain all of their limit points. Compete space: -- ...


1

You have taken the circle and decided to consider four points interchangeable. It doesn't matter what four points they are. You can think of pushing each one in to the center so they are all $(0,0)$ and you have a four petal flower. Each petal comes from one quadrant of the circle. A sketch:


2

Yes, indeed. This is the function of the second component $i$, resp. $j$. We can have all the same spaces $X_i = X$ and still have $|I|$ (where $I$ is the index set of the $X_i$) many copies of that space, all disjoint (side by side essentially, no interaction between them).


1

A compact Hausdorff space $X$ with a countable base is indeed metrisable. This is Urysohn's metrisation theorem. The standard proof embeds $X$ into the metric space $[0,1]^\mathbb{N}$. If a compact space $X$ is metrisable (with metric $d$), it has a countable base (even Lindelöf is enough for this, i.e. every open cover has a countable subcover). For this, ...


1

$$f:\mathcal{P}(\mathbb{K})\to \mathcal{P}(\mathbb{K})$$ Where $\mathcal{P}(\mathbb{K})$ denotes the power set of $\mathbb{K}$.


1

The usual definition of compactifications is actually a bit more general than that: more formally a compactification of a space $X$ is a pair $(i,Y)$, where $Y$ is compact Hausdorff and $i:X \rightarrow Y$ is an embedding (so $i: X \rightarrow i[X] \subseteq Y$ is a homeomorphism) such that $i[X]$ is dense in $Y$. Two compactifications $(i,Y)$ and $(j,Z)$ ...


2

Since compactifications also depend on how the embedding is done, we can cheat a little bit (or a lot). Consider the space $X = (0,1) \times \{ 0, 1 \}$; that is, two disjoint copies of the open unit interval. Two inequivalent but homeomorphic compactifications of $X$ are $$\begin{align} Z_1 &= ( S^1 \times \{ 0 \} ) \cup ( [0,1] \times \{ 1 \} ), \\ ...


4

No. Let $X=c_0$, so $X^*=\ell^1$. Define $T:\ell^1\to\ell^1$ by $$Tx=T(x_1,\dots)=\left(\sum_{j=1}^\infty x_j,0,0,0,\dots\right).$$ Let $V=\{x\in\ell^1:|x_1|<1\}$. If you go back to the definitions you can verify that $V$ is a weak* neighborhood of the origin but there is no weak* neighborhood of the origin mapped into $V$ by $T$. (Given $\delta>0$ ...



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