# Tag Info

7

Let $$X = \{0\} \times\mathbb{R} \cup \bigcup_{n = 1}^{\infty} \{ (x,y) : (x-n)^2 + y^2 = n^2\},$$ in the subspace topology induced by $\mathbb{R}^2$. Then $X$ is locally connected at $(0,0)$ - every $B_r((0,0)) \cap X$ for $r > 0$ is connected - but $X$ is not locally connected at any $(0,y)$ with $y \neq 0$, so $S$ is not open. Generally, $S$ being ...

6

Here's a simple counterexample. Let $X$ be $[0,1]$ with the topology that a proper subset $U\subset[0,1]$ is open iff it is a union of intervals of the form $(1/(n+1),1/n)$ for $n$ a positive integer, with the additional constraint that if $(1/2,1)\subseteq U$ then $(1/(n+1),1/n))\subseteq U$ for all but finitely many $n$. Clearly $X$ is a continuous image ...

4

In the site corresponding to a topological space, most objects have the property that all covering sieves are connected, for the reason that you explained, but there's an exception: The empty set is an object that is covered by the empty sieve. (I don't mean the sieve $\{\varnothing\}$, which also covers $\varnothing$; I mean the sieve $\varnothing$.) And ...

3

A subset being closed doesn't preclude that subset from being open. For a simple example, every discrete space is locally connected, and every subset of a discrete space—in particular the singleton sets (which are the connected components)—is both open and closed.

3

Another standard example that is even path connected but not locally connected is the comb space $$[0,1]\times\{0\} ~\cup~ \{0\}\times[0,1] ~\cup~ \bigcup_{k\ge 1} \{\tfrac1k\}\times[0,1]$$ Every neighborhood of the point $(0,1)$ contains the top end of infinitely many of the teeth, but if the neighborhood is so small that it doesn't contain the base of ...

3

There is a standard example, let define the following set: $$\Gamma:=\overline{\{(x,\sin(1/x);x\in]0,1]\}},$$ then $\Gamma$ is connected, since it is the closure of a path-connected set but not locally connected, since in any neighborhood of zero, you see disjoint line segments. Roughly, local-connectedness means that whenever you zoom in on a point, you ...

3

The comb space is an example with similar properties to the topologists sine curve... See https://en.m.wikipedia.org/wiki/Comb_space. In particular, the deleted comb space has the properties you require...

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1.) This is already in math notation (words don't make it unmathematical). If you want you can write it as $$M:=\bigcup_{q\in \mathbb{Q}\cap [0,1]} \big(\{ (t, qt)\in \mathbb{R}^2 \ : \ t \in [0,1] \} \cup \{ (t, -q + qt) \in \mathbb{R}^2 \ : \ t\in [0,1] \} \big).$$ 2.) You can connect the points $(0,0)$ and $(1,0)$ via the path $t\mapsto (t,0)$. The ...

2

Let us consider the space $\mathbb R^2$. The set $A=(\mathbb R\setminus\mathbb Q) \times\mathbb R$ is constituted of vertical parallel lines supported by irrational abscissas. This set is disconnected as you can have two open sets $]-\infty,q[\times\mathbb R\ ,\ ]q,+\infty[\times\mathbb R$ separating lines apart the rational abscissa $q$. Let's add some ...

2

Would this tell us much about the space $X$? It can tell us something about nice embeddings of $S$, and by extension, something about $X$. If $X$ is compact and $S$ is open, then shrinking $X\setminus S$ to a single point $\infty$ produces a compact locally connected space $S\cup \{\infty\}$. In particular, $X$ maps onto a locally connected space via a ...

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The conjecture is false, such a space is clearly HLC. See e.g. "Graph-Like Continua, Augmenting Arcs and Menger's Theorem."

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Some Googling finds us: this Master's thesis on the homotopy types of such continua. This paper discusses the first homology groups. this paper also seems relevant. or this Have you considered the Menger universal curve (which is a one-dimensional Peano continuum that contains a homeomorphic copy of all one-dimensional separable metric spaces. I don't ...

2

Hint: To show $\Leftarrow$ analyze what happens (looking at open basis sets) when you remove the apex $X \times \{1\}$ from the cone. Can you get results for the topological product $X \times [0,1)$ (which is embedded in $CX$)? Next, show that there are basis open sets about the apex of $CX$ that are path-connected (you get an obvious path connecting each ...

2

If I slip on this one, I’m sure others will correct me. A neighborhood of a point $P$ is just an open set containing $P$. A set $S$ of any kind is connected if it is not the disjoint union of two nonempty open subsets. A set $S$ is path connected if for any two points, $s,t\in S$, there is a continuous map $\gamma:[0,1]\to S$ with $\gamma(0)=s$ and $\gamma(... 2 Searching π-Base for (Connected + ~Path Connected + ~Locally Connected), we get: Pointed Rational Extension of$\mathbb{R}$Pointed Irrational Extension of$\mathbb{R}$Indiscrete Rational Extension of$\mathbb{R}$Indiscrete Irrational Extension of$\mathbb{R}$An Altered Long Line The Infinite Cage Smirnov's Deleted Sequence Topology Irrational Slope ... 1 We shall prove the slightly more general Theorem. For a locally compact Hausdorff space$X$the following are equivalent: (1)$X$is locally connected and has only finitey many components. (2) For each compact subset$K \subset X$and each open set$U \subset X$containing$K$, all but finitely many components of$X \setminus K$are contained in$U$. ... 1 As far as what you've done goes, I'd remark that ... since$J\cap K\cap A$happens to be empty because I said so, there is no absurd here. Rather, I think you should observe that connected components of open sets in a locally connected topological space are open, and then you should consider the connected component of$K$that contains$S$. 1 The precise statement we wish to prove is that there exists a point$p$in our space$S := A \cup B$and an open neighbourhood$U$of$p$such that for all open neighbourhoods$V$of$p$such that$V \subset U$,$V$is disconnected. Let's take$p = (0, -\frac 1 2 )$and$U = \{q \in S : \| q - p\| < \frac 1 4 \}$. Notice that$U$does not intersect the$...

1

Compactness of certain sets is not needed. For the first part of your question you will find an answer in Definition of locally pathwise connected. But for the sake of completeness let us prove once more that the following are equivalent: (1) $X$ is locally connected (locally path connected), i.e. has a base consisting of open connected (open path ...

1

Normally the subgraph induced by a set of vertices means those vertices and all edges between them that were in the original graph. So in your case it would be the four red vertices and the one edge between the topmost two of them. There is also sometimes a distinction between "open neibourhood" (all vertices adjacent to $a$) and "closed neibourhood" (all ...

1

(1) If $CX=I\times X/(0,x)\sim (0,x_0)$ is locally connected, then note that $U:=(1-\epsilon,1]\times X$ is open in $CX$. Note that topology in $U$ is corresponded to that of $X$. So since $U$ is locally connected, then so is $X$. (2) Assume that $X$ is locally connected. Then $p$ is not a vertex point in $CX$. Then there is open $U'$ containing $p$ s.t. $... 1 The intuition behind the several examples constructed for you here is as follows: take some space that you know to have many disconnected components that are 'very close' to each other. Examples include$\mathbb{Q}$and$\{0,1/n, n\geq1\}$(for which the disconnected components are the one-element subsets). These spaces are not locally connected because of ... 1 To say that$E$is the disjoint union of open connected sets is not the same as to say that$E$is the disjoint union of open balls. The idea is to first formulate a general definition and prove a general fact about any topological space. Define a component of$E$to be a maximal connected subset$C \subset E$, meaning that$C$is connected but any subset ... 1 Let$x\in E'$, where$E'$is a connected component of$E$. We know$x$is contained in a connected open subset$B_r(x)\subseteq E$. But$E'$is a maximal connected set containing$x$(since it is a connected component of$E$), so$ x\in B_r(x)\subseteq E'$. Therefore$E'$is open in$E$, since each$x\in E'$has an open neighbourhood$B_r(x)\subseteq E'$. ... 1 A nice metric example where$S$and$K$are both singletons (similar to example 119 in Counterexamples in Topology) is explained in this answer. More drastic examples, where every component of$X \setminus S$is separated from$S\$, can be found by looking for T1 spaces with a dispersion point, such as the one-point compactification of the rationals or the ...

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