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12

In $\mathbb R^2$, draw lines from $(0,0)$ to $(1,q)$ for all rational $q$ in $[0,1]$ and from $(1,0)$ to $(0,-q)$ for all rational $q$ in $[0,1].$ This subspace is path connected and nowhere locally connected.


10

It is impossible to have uncountably many equivalence classes. Note that each equivalence class is an open set, since balls are path-connected and so if $x\in G$ then any open ball around $x$ contained in $G$ is in the same equivalence class. Now any nonempty open subset of $\mathbb{R}^n$ contains an element of $\mathbb{Q}^n$, so each equivalence class ...


10

Here is an illustration of the comment above, and another example. The set $L\cap\Delta$ considered here is a subset of a line, or if we consider a parametrization, a subset of $\Bbb R$. The connected components are thus intervals. I think the main question is: how many intervals are there? First, a plot of the roots of the polynomial $(x-0.9)^3$ and of ...


8

No, a non-Hausdorff path-connected space with more than one point need not be uncountable. For a trivial example, note that if $X$ has the indiscrete topology, then any map to $X$ is continuous, so trivially any two points can be connected by a path. Less trivially, let $X=\{a,b\}$ with $\{a\}$ open (but $\{b\}$ not open), then for any open set $U\subseteq ...


8

Here's how you can get a counterexample for $n>2$ assuming the continuum hypothesis. Note that if $A\subseteq\mathbb{R}^n$ is path-connected and $x,y\in A$, there is a path-connected closed subset $B\subseteq A$ containing $x$ and $y$ (namely, the image of a path between them). So for $f$ to be path-connected, it suffices for $f(A)$ to be path-connected ...


8

You are right. The set $S_1$ is path-connected. Therefore, if $f$ is a continuous function whose domain is $S_1$, $f(S_1)$ is path-connected too. But $S_2$ is not path-connected, since $(0,\pm1)\in S_2$, but there is no element $(x,y)\in S_2$ such that $y=0$. You can use the same argument using connectedness instead of path-connectedness.


7

There are some utterly weird connected subgroups of $(R^2,+): F. B. Jones, Connected and disconnected plane sets and the functional equation $f(x)+f(y)=f(x+y)$. Bull. Amer. Math. Soc. 48, (1942) 115–120. Ryuji Maehara, On a connected dense proper subgroup of $R^2$ whose complement is dense. Proc. AMS 97 (1986) 556–558. Given such examples, I am very ...


7

Reduce the problem to showing that $\mathbb{R}^2$ minus a countable set of points is path-connected, using one of these methods. Let $S^2 \subset \mathbb{R}^n$ be a 2-sphere that contains $p$ and $q$ on the surface. A line intersects a sphere at most twice, so $C \cap S^2$ is countable. Remove any point $x \in S^2 \setminus (C \cup \{p, q\})$ and identify $...


7

I cannot speak about the whole of the complex dynamics community of course. If you are new to the notions of path-connectedness and local connectedness the Mandelbrot set might not be the best topic to gain intuition, since as you said those are current research areas and therefore well beyond the scope of a first course in topology. That said, there are ...


6

For a path between $x$ and the North Pole $\langle0,0,1\rangle$, assume that $x$ is not the South Pole. Consider $\phi:[0,1]\to S^2$ defined by the straight-line path between $x$ and the North Pole in $\Bbb R^3$ (in symbols, $\phi(t)=(1-t)x+t\langle0,0,1\rangle$). Then the projection of this path into the sphere: $$t\mapsto\frac{\phi(t)}{\|\phi(t)\|}$$ is ...


6

No, your argument needs a lot of work. Pick $x_0 \in U$ and use the hint to define $$A =\{ y \in U: \exists f: [0,1] \to U \text{ continuous and } f(0) = x_0, f(1) = y\}$$ which is all points of $U$ that can be reached by a path starting from $x_0$. Then $A$ is non-empty ($x_0 \in A$ as witnesses by a constant map from $[0,1]$ with value $x_0$). $A$ ...


6

One can find examples as subspaces of $\mathbb{R}^n$. Try taking advantage of the fact that any set between (in the sense of containment) a connected set and its closure—up to the closure itself—is also connected*. It can be the case, when $n>1$, that the closure of a path-connected (hence connected) set $X$ is no longer path-connected, which means we ...


6

I do not know how to answer your question, but I can answer it in the negative if you replace $\mathbb{R}^2$ by $\mathbb{R}^n$ for $n>2$. I will give a counterexample in $\mathbb{R}^3$; for general $n>2$ you can then just take the product of my example with $\mathbb{R}^{n-3}$. I will discuss a bit about why $n=2$ is harder at the end. Let $C\subset\...


6

Most obvious: given two arbitrary points, try to construct a path. This should work for the product problem (hint: if the factors are path connected, then this provides you with some paths to work with...) Look for counterexamples. The topologist's sine curve and its variants are the most commonly encountered counterexamples to path connectedness. This ...


6

It is not even possible for $m = n-1$. Let $E = \{ e_1,\dots,e_n \}$ be the standard basis of $\mathbb{R}^n$. Let $V$ be the the subspace generated by $\{ e_1,\dots,e_{n-1} \}$ (i.e. $V = \mathbb{R}^{n-1} \times \{ 0\}$) and $V_j = V$ for $j =1,\dots,n-1$. Let $V_n$ be the subspace generated by $\{ e_1,\dots,e_{n-2},e_n \}$. Finally let $U = E$ and $W = \{ ...


6

For open subsets $A\subseteq\mathbb{R}^n$, no choice is needed. Indeed, note that if $B\subset\mathbb{R}^n$ is any closed ball, there is a continuous surjection from an interval to $B$. Concatenating infinitely many such space-filling curves onto smaller and smaller balls around a point $a$, we get a single continuous path $\gamma:[0,1]\to A$ such that a ...


5

I assume that you mean that $A=\Bbb Q^2\cup\Bbb P^2$, where $\Bbb Q$ is the set of rational numbers, and $\Bbb P$ is the set of irrational numbers. (In other words, either both coordinates are rational, or both are irrational.) HINT: If $L$ is a line of slope $1$ or $-1$, and $L\cap\Bbb Q^2\ne\varnothing$, then $L\subseteq A$. This gets you path-...


5

Here's something that I came up with. The proposition below is what you're asking for but I've also encapsulated the main idea behind these results in the following lemma in case that's more helpful. Lemma: Let $S \subseteq X$ be path-connected and $x^1 \in \overline{S}$. Suppose there exists a countable decreasing (i.e. $U_{i+1} \subseteq U_i$) ...


5

Yes, a path-connected space $X$ consisting of more than 1 point has to be uncountable provided it is $T_1$ (you do not need Hausdorff). Otherwise the interval $[0,1]$ can be represented as the disjoint union of (more than one) countably many nonempty closed subsets, which is impossible, see this post.


5

Something which is generally true is that a homotopy equivalence $\alpha: X\to Y$ induces isomorphisms on all homotopy groups. That is, $\alpha_*: \pi_n(X) \to \pi_n(Y)$ is an isomorphism for all $n$. In the case $n=0$, $\pi_0$ is just a set, not a group, and $\alpha_*$ is a bijection. Thus $X$ is path connected iff $\#\pi_0(X) = 1 \iff \#\pi_0(Y) = 1$ iff $...


5

Yes. Clearly, any open subset of $\Bbb R^n$ is locally pathwise connected. By Exercise 6.3.10 from Engelking’s “General topology” each connected and locally pathwise connected space is pathwise connected. This result follows from the fact that in a locally pathwise connected space a maximal locally pathwise set containing a given point is open.


5

Here's a counterexample. Let $X\subset\mathbb{R}^2$ be the topologist's sine curve $\{0\}\times[-1,1]\cup\{(x,\sin(1/x)):0<x\leq 1\}$ and let $Y$ be an arc from $(0,-1)$ to $(1,\sin 1)$ that does not intersect $X$ except at the endpoints, so $X\cup Y$ is a Warsaw circle. Let $Z$ be an arc that oscillates and approaches a segment of $Y$, so the union of $...


5

Yes, indeed, a circle-like Peano continuum $X$ is homeomorphic to the circle. In order to prove this you first show that each 2-point subset $\{x, y\}$ of $X$ separates. Indeed, by taking $\epsilon>0$ small enough, for each "cyclic" cover $C_1,...,C_n$ we have $x\in C_1$, $y\in C_j$, $1\ne j$. Then no two points $z\in C_i$, $w\in C_k$, $1<i<j<k$ ...


5

You can remove fewer than $2^{\aleph_0}$ lines from $\mathbb R^3$ and the remainder $X$ will still be path-connected. Consider any two points $P,Q\in X.$ Consider the sphere having $PQ$ as a diameter. Each of the missing lines meets the sphere in at most two points; thus there are fewer than $2^{\aleph_0}$ missing points on the sphere. The $2^{\aleph_0}$ ...


5

I think the answer is no. Take for instance the subspace $X$ of $\mathbb{R}^2$ defined by $$ X = \{(x,\sin 1/x): x > 0\} \cup \{(0,y) : -1 \leq y \leq 1\}. $$ This space has exactly two path components $$ A = \{(x,\sin 1/x): x > 0\} \text{ and } B = \{(0,y) : -1 \leq y \leq 1\}. $$ So $X/\sim = \{A,B\}$. Let $\pi: X \to X/\sim$ be the quotient map. ...


5

Perhaps that it can work, but that approach looks too much complicated. Besides, there are $2$-dimensional manifolds which cannot be embedded in $\mathbb{R}^3$ (the Klein bottle is an example). The statement is easy to prove using the fact that manifolds are locally homeomorphic to $\mathbb{R}^n$, together with the fact that, in $\mathbb{R}^n$, open ...


5

If $\gamma_2(t_0) \le 0$ then the above argument works. If $\gamma_2(t_0) \ge 0$, then replace $\frac{2}{4n+1}$ everywhere in the argument with $\frac{2}{4n-1}$ so that $\gamma_2(t) = -1$. In either case you are able to find a $t \in (t_0, t_0 + \delta)$ such that $|\gamma_2(t) - \gamma_2(t_0)| \ge 1$.


5

We have to assume that $M$ is open or that every open ball in $M$ is path connected in $M$. Otherwise, as the OP himself pointed out, the topologist's sine curve provides a counterexample. Assuming that the hypothesis is modified here is a hint: let $x \in M$. Let $S$ be the set of all points $y \in M$ such that there is a path in $M$ from $x$ to $y$. It is ...


5

Consider $X = \Bbb N$ in the cofinite topology, then $X$ is connected but any continuous $p: [0,1] \to X$ is constant. As noted in the comments, Hausdorff examples also exist, but we cannot get regular Hausdorff countable connected spaces (as then maps onto $[0,1]$ exist by normality).


5

First, note that this formula only applies to (in fact only makes sense for) line integrals of certain vector fields $\bf F$, namely conservative ones, that is, those of the form $${\bf X} = \operatorname{grad} F$$ for some function $F : U \to \Bbb R$ and (when $n > 1$) this is a very strong restriction: In a sense that can be made precise, most vector ...


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