New answers tagged general-topology
5
votes
Accepted
Is there a continuous bijective mapping of $R$ into a compactum
We can modify the example of the continuous bijection $[0, 1) \to S^1$ slightly; namely, thinking of $\mathbb{R}$ as an open interval $(0, 1)$, it admits a continuous bijection to a compact subspace ...
- 432k
2
votes
When will "Retract $\iff$ Deformation Retract" hold true?
Statement 1 is almost never true. For example if $A$ is a point it is always a retract of $X$ (via the unique map $X \to A$) but if it's a deformation retract then $X$ must be contractible. Being a ...
- 432k
2
votes
Examples of topological spaces that closed+boundness implies compactness
Here is a criterion similar to what you asked for in the question. If $(X,d)$ is a metric space, we say that $E\subseteq X$ is totally bounded if for every $\varepsilon>0$, there exists a finite ...
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1
vote
Examples of topological spaces that closed+boundness implies compactness
In locally convex theory, Montel's theorem was the reason to call barrelled spaces with the property that all closed bounded sets are compact Montel spaces. As a rule of thumb, almost all locally ...
- 12.3k
0
votes
Homotopy equivalence of the tetrahedron minus its interior
Contract the base triangle to a point. This gives you two vertices (the top vertex and what used to be the base) connected by three separate arcs. Contract one of the arcs, and you get the wedge of ...
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2
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Closed sets definition in complex analysis
Neither is more correct since we can prove:
A set $A \subset \mathbb{C}$ contains all its limit points if and only if it contains all its boundary points.
However, the first definition is more ...
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Is n-euclidean space a quotient of m-euclidean space?
Assume that we have a quotient map from $\mathbb{R}$ to $\mathbb{R}^2$. Then its square is a quotient map from $\mathbb{R}^2$ to $\mathbb{R}^4$ ( product of quotient maps OK for locally compact), so ...
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0
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Examples of topological spaces that closed+boundness implies compactness
Think of this answer more as a comment-answer that would be a bit too long to fit into a comment. I'm sure there are more knowledgeable people who would be able to answer with more confidence and more ...
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0
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Proper direct image and extension by zero
In the context of locally compact Hausdorff space, all notions of topological properness are the same. However, if you're doing sheaf theory and, say, you want to do Verdier duality in general, then ...
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10
votes
Accepted
Is n-euclidean space a quotient of m-euclidean space?
Does there exist a quotient map $f : \mathbb{R}^m \to \mathbb{R}^n$?
Yes. Even a closed map. Even when $m>n$ (note that while projections are quotient maps, they are not closed).
Consider $k\in\...
- 44.6k
0
votes
$f$ is open iff $f(Int(A)) \subset Int(f(A))$
Suppose $f$ is open. Let $A \subset X$.
Since $\text{Int}(A)$ is open in $X$ and $f$ is an open map, $f(\text{Int}(A))$ is open in $Y$. Since $\text{Int}(A) \subset A, f(\text{Int}(A)) \subset f(A)$. ...
1
vote
Accepted
Every disjoint union $A\sqcup A$ of two same connected sets is disconnected?
The topology on $X\sqcup Y$ is defined as $A$ is open iff $A\cap X$ is open in $X$ and $A\cap Y$ is open in $Y$.
Thus, $U$ is open in $A\sqcup A=A\times\{0,1\}$ iff $U\cap(A\times\{0\})$ is open in $A\...
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1
vote
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Convergence of Sequences with respect to Ultrafilters on the Natural numbers in Compact Hausdorff Topological Spaces
For part $1$, assume $X$ is compact. Assume to the contrary that $\lim_\limits{\mathcal{F}} (x_n)$ is empty. Then by definition of convergence w.r.t. $\mathcal{F}$, we see that for any $x \in X$, ...
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5
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Does a continuous surjective map on topological spaces always have a continuous right-inverse?
This is very, very far from true, for quite interesting reasons. Here are some suggestive examples.
Let $f : Y \to X$ be a covering map between reasonable path-connected spaces, classified by a ...
- 432k
0
votes
Open Equivalence relation induces discrete topology on the quotient
My idea is to show that maybe $\pi^{-1}(\{[x]\})=U$
That would be nice, but note that the introduction of $U$ involved an arbitrary choice. You can't hope for that to be true independent of the ...
- 6,545
0
votes
Proving that the set of limit points of a set is closed
Let $x \in E'$ be a limit point of $E'$.
$\implies \forall$ neighborhoods $U$ of $x$ $ \exists z \in E'$ such that $z \in U$.
Since, $U$ is also a neighborhood of $z$ and $z \in E' \implies \exists y ...
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3
votes
Name for topology generated by the standard open sets of reals and every point that isn't zero.
Expanding on the answer of @StevenClontz, the following more general characterization is true.
Suppose $X$ is a topological space such that
every point of $X$ is isolated, expect for one special ...
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6
votes
Accepted
Confusing Remark on the Deformation Retraction from $\mathbf{X}$ to $\mathsf{X}?$ - Hatcher's Algebraic Topology
As defined about a page before the part you quoted,
A deformation retraction of a space $X$ onto a subspace $A$ is a family of maps $f_t : X \to X$, $t \in I$, such that $f_0 = 1$ (the identity map), ...
- 124k
0
votes
A function with zero gradient is locally constant
Proof (A):
Assume there are $v,u\in X$ s.t. $f(v)\neq f(u)$. Define $g:\mathbb{R}\rightarrow\mathbb{R}$ by $t \mapsto f(u+t(v-u))$. We have
$$
g(0) = f(u)\neq f(v) = g(1)
$$
by assumption. Also,
$$
g'(...
- 53
1
vote
Group actions on ringed topological spaces
For any map of ringed spaces $f:X\to Y$ and any open set $W\subset Y$, we have an induced map $\mathcal{O}_Y(W)\to \mathcal{O}_X(f^{-1}(W))$ (ref Stacks 0090). Now apply this to the situation where ...
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8
votes
Is a continuous function a generalization of an adjunction?
The author is alluding to an extremely general construction called the Chu construction which generalizes adjunctions of categories as well as continuous functions, and a bunch of other things. I will ...
- 432k
4
votes
How to Prove a Set is Closed in $\mathbb{R}^2$ using the product topology definition?
If you want to go down the rabbit hole, then you really should explain why
there exists an $\epsilon>0$ such that for all $(x,y)\in (a-\epsilon,a+\epsilon)\times(b-\epsilon,b+\epsilon)$, we have $...
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3
votes
Accepted
Well-definedness of the projection associated to the sheaf of germs of a presheaf
$F_x$ is defined to be a set depending on $x$, and $F$ is defined to be a disjoint union of these sets.
So purely as a matter of definition, if $x\neq y$ then $F_x\cap F_y=\emptyset$, and likewise $[s]...
- 30.3k
1
vote
Accepted
Continuous injection to manifold with boundary
From the fact that $U \subset M$ is open and that $f : U \to M$ is continuous, injective, and open, it follows that $V=f(U)$ is open in $M$ and that $f$ restricts to a homeomorphism $f : U \to V$. ...
- 124k
3
votes
Accepted
Must scattered spaces with points $G_\delta$ be locally countable?
Consider $\mathbb{R}$ with the following topology: a set $U$ is open iff either $0\not\in U$ or $U$ contains an open interval around $0$. This topology is scattered (every point except $0$ is ...
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0
votes
Trouble Understanding Baby Rudin $2.38$
Let $a_n$ and $b_n$ be the lower and upper bounds of each interval $I_n$ respectively. Then clearly $a_n \leq b_n$ fpr each interval. Thus $\sup{a_n} \leq \inf{b_n}$ given that $I_{n} \subset I_{n+1}$....
1
vote
Accepted
Specific construction of loop in Union of path connected spaces
$D$ being clopen implies it is all of $(0, 1]$ since $(0, 1]$ is connected, and $D = (0, 1]$ is in particular to say that the statement holds for $\delta = 1$ which is what you're trying to show.
As ...
- 4,963
0
votes
Accepted
Proof of $f(\overline{S}) = \overline{f(S)}$
Note that, a map $g:X \to Y$ is continuous if and only if $g(\overline{E}) \subseteq \overline{g(E)}$ for any subset $E \subseteq X$. You can see a proof of this here.
With this in mind, note that, by ...
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4
votes
Accepted
Is a topological abelian group with no open subgroup connected?
If you add a compactness condition, then the result is true.
Consider a topological group $G$ with no proper open subgroup (I am not assuming $G$ to be abelian). The connected component of the ...
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3
votes
Understanding proof of existence of Schreier transversals
Your question is answered by a standard application of Van Kampen's Theorem, and the set $J$ is defined in the second sentence of the second paragraph of the proof. Perhaps it will be clearer if one ...
- 124k
8
votes
Accepted
Is this "continuous" function really continuous?
No, this is not necessarily true. As an example, let $n = 2$, $\phi_t: \mathbb{R}^2 \to \mathbb{R}^2$ be defined by,
$$\phi_t(x, y) = \begin{cases}
(t^{|x|}, y) &, \text{ if }t \in (0, 1], x \in [-...
- 9,938
0
votes
Why does an open map send generic points to generic point?
Here is supplment for the Eric Wofsey's answer, for more clarity. Let $f :X \to Y$ be a open morphism locally of finite type to a irreducible noetherian scheme $Y = \overline{ \{ \eta\}}$. We want to ...
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0
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What is the uniform metric on $\mathbb{R}^X$?
The uniform metric is a function $\bar{\rho}: \mathbb{R}^X \times \mathbb{R}^X \to \mathbb{R}$ defined as
\begin{equation}
\bar{\rho}(f,g) = \sup \cup_{x \in X} \{|f(x)-g(x)|\}
\end{equation}
where $f,...
1
vote
Accepted
Intermediate Value Theorem on a Tree
The answers to your questions are both yes. For any tree $T$, any two points $x \ne y \in |T|$, and any subspace $A \subset T$, let's say that $A$ is a subarc with endpoints $x,y$ if there exists a ...
- 124k
1
vote
In a regular space with a $\sigma$-locally finite network, is every closed set a $G_\delta$?
Details on some lemmas that @Ulli used implicitly, shown here for convenience:
Lemma 1. If $\mathcal N$ is locally finite, then $\bigcup\{\overline N:N\in\mathcal N\}=\overline{\bigcup\mathcal N}$.
...
0
votes
The Closure of a Connected Set is Connected
But If $A \cap V = \varnothing$, as $\{a\} \notin U \cap V$, then, If assume $a \in V$ without loss of generality, $C = (A \cap U) \cup (\{a\} \cap U) = \varnothing$. Therefore, we have contradict.
2
votes
Accepted
In a regular space with a $\sigma$-locally finite network, is every closed set a $G_\delta$?
Let $\mathcal N = \bigcup_{n \in \mathbb N} \mathcal N_n$ be a network, such that each $\mathcal N_n$ is locally finite.
Then for each n, $\{\overline{N}: N \in \mathcal N_n\} $ is locally finite, ...
- 4,307
2
votes
Accepted
Equivalence of two strengthenings of open neighborhood
The other implication seems to be false. For instance, take $X = \lbrace \frac{1}{n} : n = 2,3,4,\ldots\rbrace$ and $N = ]0,1[$ subsets of $\mathbb R$. Then $N$ is open, but does not contain $0\in \...
- 516
1
vote
Accepted
Verifying proof that in a first countable space if $u_n\to u$ and $f(u_n)\to f(u)$, then $f$ is continuous
The fact that $u \in \text{cl}(A)$ implies that there is a sequence $\{u_n\}$ of points in $A$ that converges to $u$ is true because the space is first countable.
This is not true for any topological ...
- 21.2k
2
votes
Accepted
If $M$ is a connected topological manifold and $p,q \in M^{o}$, then there exists a homeomorphism which maps $p$ to $q$.
I'm not sure how 'explicit' you can get the homeomorphism but this is the general construction. Given a path from $p$ to $q$ in $M$, for sufficiently small $\varepsilon$ the $\varepsilon$-neighborhood ...
- 6,113
0
votes
Can a function over a domain with holes have a derivative of zero everywhere but fail to be constant?
The problem in your logic is that "continuous functions with a derivative of zero everywhere must be constant". This is only true in connected domains. For example, in $(- \infty, 0)$, you ...
- 91
1
vote
Accepted
Compactess of a set in $\mathbb{R}^d$ defined as the union of compact sets
The Heine-Borel theorem tells us that it is sufficient to check that $\mathcal{A}$ is closed and bounded. Rewrite $\mathcal{A}$ as:
$$
\mathcal{A} = \{y \in \mathbb{R}^d: \exists x \in [a, b], \...
- 2,252
1
vote
Accepted
Non-proper smooth embedding
No: the embedding $(-1,1) \to \Bbb R^2$ given by $t \mapsto (t,0)$ is smooth, but not proper, since the preimage of the compact subset $[-1,1]\times\{0\}$ is not compact.
- 20k
1
vote
Accepted
Composition of asymmetric contraction mappings
Yes: $d(J(T(m_1)), J(T(m_2))) \leq kq(T(m_1), T(m_2)) \leq kcd(m_1, m_2)$.
- 1,579
1
vote
Accepted
Is every compact set equal to an intersection of nested bounded sets with smooth boundary?
Here is the standard procedure for constructing such open subsets in $\mathbb R^n$. Let $C\subset \mathbb R^n$ be any closed subset. For $r>0$ define its closed $r$-neighborhood
$$
\bar{N}_r(C):=\{...
- 102k
4
votes
Connectedness of a subspace of $M_2(\mathbb C)$
Your space is homeomorphic to $\Bbb C^*\times\Bbb C^3$. A product of connected spaces is connected.
- 39.7k
1
vote
Accepted
Locally compact metric space and separation property
Yes. Take $N = X \setminus U_2$. We have $U_1 \subset N$, thus $N$ is a closed neighborhood of $C$. The interior of $N$ is $\operatorname{int} N = X \setminus \overline U_2$ and the boundary of $N$ is ...
- 78.2k
3
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equality involving supremum and infimum of two disjoint sets that partition the real Line
You're having trouble with this because it's false. Let $B=\{0 \} \cup (1, \infty).$ Let $A= \Bbb R \setminus B$. Then $\sup A=1$ but $\inf B = 0$. For this to be true, you need the additional ...
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0
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Are functions from a non metrizable general topological space into $\mathbb{R}$ continous under the ring structure?
Alright, from the hints I got what to do. For those with the same question, here is the solution:
Consider the following commutative diagram of topological spaces:
[]
where,
$i: x \to (x,x)$
$p: (a,...
1
vote
Accepted
What is the meaning of a "regular open basis" for a topological space?
For your second question, consider the half-open interval topology on the real line, generated by the sets of the form $(-\infty, a)$ for $a \in \mathbb{R}$. The closure of each nonempty open set is ...
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