New answers tagged

0

By simple computations, one checks $AB=BA$, $AC=CA$, $CB=BCA$ is a presentation. So in the abelianization $A=1$, $BC=CB$ the abelianized group is $\bf Z^2$. Note that $dy, dz$ are invariant by $G$ and form a base for the de Rham co-homology


1

Getting a finite open cover $K$ consisting of metric balls seems to be a red herring. Your assumptions imply there is an open $U$ containing $K$ over which $f$ is injective. Let $\epsilon=d(K,\mathbb R ^n \setminus U)$. All you have to show is $\epsilon>0$, because clearly $\{x\in \mathbb R^n : d(x,K) < \epsilon \}\subseteq U$. Suppose ...


0

Here’s another variation on the idea of using compactifications to avoid algebraic topology: $\Bbb R\times[0,1]$ has a two-point compactification, but $\Bbb R^2$ does not. The two-point compactification of $\Bbb R\times[0,1]$ is pretty evident. Suppose that that $\Bbb R^2$ has a compactification $X=\{p,q\}\cup\Bbb R^2$, where $p\ne q$. Let $U$ and $V$ be ...


1

In a Hausdorff space, compact sets are closed. So $K_1\cap K_2$ is closed in $K_1$. Since a closed subset of a compact set is compact, you're finished.


2

The proof as it stands is fine. This method is the "index-juggling" method, so to say. You could also use a more "point-centric" approach, which comes down to the same: suppose $O$ is open in the product topology on $X \times Y$. Suppose $(p,q) \in O$. Then there are open sets $O_p \subseteq X$ and $O_q \subseteq Y$ such that $(p,q) \in O_p \times O_q ...


0

Proof without nets: Let $G$ be the graph of $f$. Suppose $G$ is compact. Let $A$ be closed in $\mathbb R$. Consider $G\cap (X\times A)$; this set is compact. Note that $f^{-1}[A]=\pi_X [G\cap (X\times A)]$. The projection map $\pi _X$ is continuous, so it maps compact sets to compact sets. As a compact subset of a Hausdorff space ($X$) is closed, ...


-2

You need to show that for every $x\in M$ you can find an open ball that is still in $M$. The open Ball definition is : \begin{align} B_r(x) = \{ y\in \mathbb{R^3} | \sqrt{\sum_{i=1}^3 (y_i-x_i)^2} < r)\} \end{align} So you have an inequality for all y in the ball. You have an inequality for all x int the set $M$. Combining the two if you can find a $r$ ...


2

Let $f(x_1,x_2,x_3)= x_1^2-2(x_2^3+x_3^3)$, which is a polynomial, hence continuous. Then the set in question is $f^{-1}([0,\infty))$ which is closed, hence by the definition of continuity, the inverse image of this closed set is closed.


3

Yes, it is separable - in fact, it has a countable dense subset consisting entirely of finite sets! Exercise. Let $C$ be a closed bounded set, and $N\subseteq C$ such that for each $c\in C$, there is an $n\in N$ such that $d(n, c)<\epsilon$. Then the Hausdorff distance between $N$ and $C$ is $<\epsilon$. Such an $N$ is sometimes called an ...


1

Since $\pi$ is locally trivial, it is an open map. Thus the result follows by http://stacks.math.columbia.edu/tag/004Z which states: Let $f : X \to Y$ be a continuous map of topological spaces. Assume that (a) $Y$ is irreducible, (b) $f$ is open, and (c) there exists a dense collection of points $y \in Y$ such that $f^{-1}(y)$ is irreducible. Then $X$ is ...


0

Be careful, we still have the second criterion and must use the definition of dictionary order. That is given an order relation $<_A$ on the set $A$ and an order relation $<_A$ on the set $B$, we define the order relation $<$ on $A\times B$ such that: \begin{equation} a_1\times b_1 < a_2 \times b_2 \end{equation} if $a_1<_Aa_2$, or if ...


0

There are three different type of connectedness - you have only mentioned 2, but I think it is clearer if I state all three. The topological connectedness. This is the weakest. A stronger one is path connectedness. ~ "Any two points can be connected by a continuous function." Finally the strongest one is "polygonal-connectedness". ~ "Any two points can ...


1

I think part of your misunderstanding comes from thinking of the subspace as a subspace (remaining within the larger space) vs the subspace as its own sort of space. You are right to conclude that the disjoint union of two closed balls that do not intersect is not connected. However, a single closed ball is connected: you must think of it as its own space. ...


0

"Let B be a subset of the complex plane that can be expressed as the union of two disjoint closed ε-neighborhoods about two different points in the complex plane. " $B$ can be written as the union of two disjoint nonempty open sets in the subspace topology.


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In $\;\Bbb C\;$ and other spaces, both definitions are almost equivalent: a space is connected if it is path-connected, which means any two points can be joined by. In your examples, $\;A\;$ is path connected, whereas $\;B\;$ isn't, no matter what definition you choose. What you say about subsets unionized into...is a little beyond my comprehension.


2

Consider the letter "X" (i.e., two crossing line-segments). It's starlit from its center, but every point except the crossing-point is not a star center.


0

Here is a proof that if $n<m$ are natural numbers then every function $f : I_m \to I_n$ is non-injective. The proof is by induction on $n$. Consider first the base case $n=1$, and so $m \ge 2$. Then $f(1)=f(2)=1$ and so $f$ is not injective. Now assume the induction hypothesis: the statement is true for $n=k$. We must prove the statement is true for ...


5

The closed sets are $\{\emptyset,X,\{a\},\{b,c\}\}$, so the intersection of the closed sets containing $A$ is $\{b,c\}$. The fact that $\{b,c\}$ is also open is irrelevant: nowhere it is required that a closed set is not open. Subsets of $X$ can be open and closed open and not closed not open and closed neither open nor closed The subsets of your three ...


0

What have you tried so far? And, as for being one-to-one, can you think of examples where a function maps, say, every open interval to $R$? what would such a function look like? is it one-one? onto?


0

Not a direct answer to your question, but here is essentially your proof written down in an alternative style, which might be helpful. (See, e.g., EWD1300 for more details on this proof style.)$ \newcommand{\calc}{\begin{align} \quad &} \newcommand{\op}[1]{\\ #1 \quad & \quad \unicode{x201c}} \newcommand{\hints}[1]{\mbox{#1} \\ \quad & \quad ...


2

This result is true under weaker hypothesis - connectedness (not necessarily path connectedness) of $Y$ is all we need. Fix a point $y_1\in Y$. Let $|p^{-1}(y_1)|=k$. Suppose there is some $y_2\in Y$ such that $|p^{-1}(y_2)|\neq k$ Define the set $A:=\{y\in Y\ |\ |p^{-1}(y)|=k\}$. We would like $A$ to be all of $Y$. For this we will use the connectedness ...


0

HINT: for the first question you can use the stereographic projection to see that a sphere minus a point is homeomorphic to a plane, in particular all homotopy groups are the same. For the second question: use the Van-Kampen theorem to see that $\pi_1(S)$ is trivial. The conclusion follows.


1

Fix $x_0 \in X$ and let $x_a$ be a net converging to $x_0$ in $X$. You need to show $f(x_a)\rightarrow f(x_0)$ in $\mathbb{R}$. Let $G_f$ denote the graph of $f$, which is a subset of $X \times \mathbb{R}$. Then $(x_a,f(x_a))$ is a net in $G_f$, which is compact, so we can extract a convergent subnet $(x_b,f(x_b)) \rightarrow (x,y) \in G_f$. But then $x_b ...


1

First of all note that any two distinct points in $X$ can be separated by open sets. This is shown as follows - Let $y,z\in X$ with $y\neq z$. Then $d(y,z)=:r>0$. Then the open balls $B(y,r/2)$ and $B(z.r/2)$ are disjoint open sets where the first one contains $y$ and the second contains $z$. Now coming to your problem - For each $a\in A$ there is a ...


1

There are three different definitions of subnet in fairly common use; their definitions are given in this question and answer. I suspect that you’re using one of the definitions given in the question; the hint below is written for the second of those but is easily adapted if you’re using one of the other definitions. HINT: Let $\nu=\langle D,\le\rangle$ be ...


0

It need not be an annulus; here’s a counterexample. Let $X=\Bbb N^{\Bbb N}$, where $\Bbb N$ has the discrete topology; $X$ is homeomorphic to the irrationals. For distinct $x=\langle x_n:n\in\Bbb N\rangle$ and $\langle y_n:n\in\Bbb N\rangle$ in $X$ let $$\delta(x,y)=\min\{n\in\Bbb N:x_n\ne y_n\}\;,$$ and let $$d:X\times X\to\Bbb R:\langle ...


2

Hint: By the Intermediate Value Theorem, $f$ must be order preserving (strictly increasing). If $n>1$ and $f^n=f$, then you have $f^{n-2}\circ f=f^{n-1}=\iota$; thus the identity is the composition of two order preserving functions one of which is $f$. Now can you prove $f=\iota$?


1

Hint - Using the universal property of quotient spaces you can get a continuous map $\phi:(X\times X)/ {\sim'}\to X/ {\sim}\times X/ {\sim}$ which in this case you can show to be bijective and open.


2

Hint The homeomorphism is given by $$f:(X\times X)/{{\sim}'}\longrightarrow (X/{\sim})\times (X/{\sim})$$ defined by $$f([(x,y)])=([x],[y]).$$ Prove it !


0

Here is a proof: Let $\gamma$ be a path from $y_1$ to $y_2$. For each point $x \in p^{-1}(y_1)$, there is a unique lift $\gamma_x$ of $\gamma$ to a path which begins and $x$ and ends at a point $x'$ of $p^{-1}(y_2)$. The map $x \mapsto x'$ from $p^{-1}(y_1)$ to $p^{-1}(y_2)$ is an injection: two of these lifts can't intersect, and therefore can't end at ...


7

Any countable subset of $\mathbb{R}^2$ (with at least two elements :P) is indeed disconnected; but I think it takes a different argument than what you sketch to show this. Here's a proof outline. Let $A\subseteq\mathbb{R}^2$ be countable and fix distinct $a, a'\in A$. Say that a positive real $r$ is good if $r<d(a, a')$, and for no $b\in A$ do we have ...


3

Perhaps I will write something wrong? (This seems easy to me). First of all, I think you mean there can't exist an injective function $f:I_m\to I_n$, for if $n<m$ there actually exists an injective $I_n\to I_m$ (the inclusion). Suppose $n<m$ and $f:I_m\to I_n$ injective. Then $f:I_m\to f(I_m)$ is bijective, where $f(I_m)=\{f(1),...,f(m)\}$. Clearly ...


0

Suppose $\;X\;$ is infinite, then $\;\{\{x\}:\;x\in X\}\;$ is an infinite open cover of $\;X\;$ which has no finite subcover, contradiction.


1

Hint: Fix any $x_0$ in your domain and look at the set of $x$ in your domain for which $f(x) = f(x_0)$. Is this set open? Is it closed? What does that tell you?


1

Take the open unit ball $B$ in an infinite dimensional Banach space. If $B$ is totally bounded so is its closure (exercise!). Hence $\textrm{cl}(B)$ is complete and totally bounded, thus compact. But this is false.


4

Start with an arbitrary metric space (with metric $d$) that is not totally bounded, and take the new metric $\overline{d}(x,y) = \min(1, d(x,y))$.


3

Let $\mathscr T$ be a topology on a non-empty set $X$ and let $\mathscr S\subseteq \mathscr T$. Then, by definition, $\mathscr S$ is a subbasis for the topology $\mathscr T$ if for any $U\subseteq X$, one has that $U\in\mathscr T$ if and only if $U$ can be expressed as a union of sets which are finite intersections from sets in $\mathscr S$. Formally: ...


1

Here's a partial result. In particular, this answers your question in the affirmative for Hausdorff spaces assuming CH, or more generally for Hausdorff spaces if you replace $\aleph_1$ by $\mathfrak{c}$. Theorem: Let $X$ be a Hausdorff space with $|X|>\mathfrak{c}$. Then there is a subspace $Y\subset X$ of cardinality $\aleph_1$ which is not ...


2

A connected set is a set that cannot be divided into two disjoint nonempty open (or closed) sets. Intuitively, it means a set is 'can be travelled' (not to be confused with path connected, which is a stronger property of a topological space - every two points are connected by a curve). A simply connected set (let me short it to SC for now) is path-connected ...


1

A connected set is a set that cannot be split up into two disjoint open subsets (this of course depends on the topology the set has; for the case of $\mathbb{C}$, this is the same as the Euclidean topology on $\mathbb{R}^2$). Now, a simply connected set is a path-connected set (any two point can be joined by a continuous curve) where any closed path (a ...


1

A connected set is a set which cannot be written as the union of two non-empty separated sets. This is when the set is made only of one-part, if one wants to think of it intuitively. However, simple-connectedness is a stronger condition. It requires that every closed path be able to get shrunk into a single point (continuously) and that the set be ...


0

One would definitely need some order structure, otherwise the very notion of “sides” fails to make sense. Here is an idea. Let $X$ and $Y$ be two topological spaces and let $\succsim$ be a (say) partial order on $X$. Take any function $f:X\to Y$ and fix $x_0\in X$ and $y_0\in Y$. Let $$L_0\equiv\{x\in X\,|\,x_0\succsim x\}$$ be the “lower contour set” ...


1

I propose a proof in the spirit of what you have attempted. Let us define $$g((x_1, x_2, \dots, x_{n+1})) := x_1^2 + \dots + x_{n+1}^2 -1 $$ then $$M=g^{-1}((-1,1))$$ Thus $M$ is the reciprocal set of the open set $\mathbb{R}$, therefore an open set of $\mathbb{R}^n$.


1

We have the following inverse image $$ M^c=\|\cdot\|^{-1}\left([1,\infty) \right) $$ the subset $[1,+\infty)$ is closed in $\mathbb{R}$ thus the subset $M^c$ is closed in the metric space $\mathbb{R}^{n+1}$ ($\|\cdot\|$ is a continuous function over $\mathbb{R}$), that is $M=\left(M^c \right)^c$ is an open subset in $\mathbb{R}^{n+1}.$


1

Consider the complement of $M$ and take a sequence there to show that it is closed. Since we are working in a metric space, this approach works.


1

a) The maps $p:X\to Y$ both open and locally injective you are looking for are, practically tautologically, the local homeomorphisms. An alternative terminology (especially for us ze French) is that $X$ is an étalé space over $Y$. They are very important and correspond to the historical definition of sheaves, before the definition of sheaves as functors ...


3

Sure. Let $x\in M$. Put $\delta=1-\|x\|$. If we show that the ball of radius $\delta$ around $x$ is contained in $M,$ this implies that $M$ is open. The key fact is that $\|z\|=(z_1^2 +\ldots+z_{n+1}^2)^{1/2}$ is a norm. In particular, it satisfies the triangle inequality. So, if $\|z-x\|<\delta$, then ...


3

For "image" this is clearly impossible, since every single point is a closed subset, its image must be a single point, and those are not open. For inverse image, constant maps have this property because the preimage of anything is either the empty set or all of $\mathbb R$, and both are closed and open at the same time. The only interesting question is, ...


1

Let $f$ be an element of $C_0$, for every $c>0$, there exists $n_0$ such that $\mid x\mid>n_0$ implies that $\mid f(x)\mid <c/4$. Stone-Weirstrass applied to $C([-n_0,n_0])$ implies that there exists a polynomial $p$ such that $\|f(x)-p(x)\|<c/4$ for $\mid x\mid<n_0$. You have $p=e^{-x^2}(e^{x^2}p)$. By multiplying the analytic development of ...


0

Interior point: Let S be a set in R .x €S is said to be interior point of A if there exists e>0 such that B (x;e)subset of s



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