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This is not actually a question asked by me. But since I do not know the answer, I would love to know if someone here could answer it.

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By "subspace", I expect that you mean "subset". I edit you title accordingly. –  Denis Serre Nov 20 '10 at 17:06
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What's wrong with subspace? A subspace means a subset with the subspace topology it inherits from the ambient space. –  Todd Trimble Nov 20 '10 at 17:17
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Some people might misread as "vector subspace". –  Nate Eldredge Nov 20 '10 at 17:20
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That would be a very strange misreading! –  Todd Trimble Nov 20 '10 at 17:32
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I've searched through all the reference tables in Counterexamples in Topology. I didn't see any answers to the question. –  Todd Trimble Nov 20 '10 at 17:34
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No. There are locally connected subsets of $\mathbb{R}^2$ which are totally path disconnected. See my answer to this old MO question "Can you explicitly write R2 as a disjoint union of two totally path disconnected sets?". Also, Gerald Edgar's response to the same question says that such sets cannot be totally disconnected, although he does not mention local connectedness. In fact, the sets given by my answer are locally connected, so provide a counterexample to your question.

As in the linked question: Let $S$ be a subset of the reals such that $S\cap[a,b]$ and $S^{\rm c}\cap[a,b]$ cannot be written as a countable union of closed sets for any $a < b$ (e.g., this explicit example of a non-Borel set). Then, $A=(S\times\mathbb{Q})\cup(S^{\rm c}\times\mathbb{Q}^{\rm c})$ and $B=(S\times\mathbb{Q}^{\rm c})\cup(S^{\rm c}\times\mathbb{Q})$ partition the plane into a pair of locally connected and totally pathwise disconnected sets.

That they are totally pathwise disconnected is proven in my answer to the linked question. Let us show that $A\cap U$ is connected for any nonempty 'open rectangle' $U=(x_0,x_1)\times(y_0,y_1)$. If not, there would be nonempty disjoint open sets $V,W\subset U$ with $A\cap(V\cup W)=A\cap U$. If $\pi(x,y)=x$ is the projection onto the x-axis then $\pi(V)\cup\pi(W)=\pi(V\cup W)=(x_0,x_1)$ is connected. So, we can find a nontrivial closed interval $[a,b]\subseteq\pi(V)\cap\pi(W)$. Now, for every $x\in S^{\rm c}\cap[a,b]$, the line segment $\lbrace x\rbrace\times(y_0,y_1)$ intersects with both $V$ and $W$ and, by connectedness of line segments, it will intersect with $U\setminus(V\cup W)$. Hence, there is a $q\in\mathbb{R}$ with $(x,q)\in U\setminus(V\cup W)\subset B$. So, $q\in\mathbb{Q}$. For each rational $q$, let $S_q$ be the (closed) set of $x\in[a,b]$ such that $(x,q)$ is in $U\setminus(V\cup W)$. Then, $S^{\rm c}\cap[a,b]=\bigcup_{q\in\mathbb{Q}}S_q$ is a countable union of closed sets, giving the required contradiction. Hence $A$ is locally connected and, exchanging $S$ and $S^{\rm c}$ in the argument above, so is $B$.

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Great answer! But I think it would be easier to read if you used TeX instead of HTML for the math. For one thing, having variable names appear in italics helps a lot. –  Nate Eldredge Nov 20 '10 at 19:02
    
Ok, since you ask, I'll change it to use latex. In the early days of MO, latex support was not so good, so I got into the habit of using html (and for long answers, latex can still be cumbersome, but that doesn't apply here). –  George Lowther Nov 20 '10 at 19:08
    
Wow, very impressive! –  Todd Trimble Nov 20 '10 at 19:10
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Isn't every connected subset of ${\mathbb R}^n$ locally a Peano space? Then the answer would be "yes".

Edit I assumed the subspace to be closed. Then it would be locally compact, hence locally Peano. I guess the main difficulty is for not closed subsets.

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What about compactness? –  Vít Tuček Nov 20 '10 at 17:08
    
It is not in general, since a Peano space is compact, by definition. –  Denis Serre Nov 20 '10 at 17:10
    
I found a proof of the fact that a Peano space is arcwise connected at books.google.com/… –  Nate Eldredge Nov 20 '10 at 18:19
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An easy counterexample: The basic space that is connected but not locally connected is the closure of the graph $y=\sin(1/x)$ ("the topologist sine curve"). Call it $S$. Pick a countable set $D$ on the graph $y=\sin(1/x)$ whose closure is $D\cup \{0\}\times [-1,1]$. For each point $z \in D$ add a copy $S_z$ of $S$ (in $R^3$) starting from $z$ and ending on an interval in $0 \times [-1,1]$ so that the diameter of $S_z$ is at most twice the distance from $z$ to the $y$-axis and $S_z$ intersects $S_w$ only along the $y$-axis if $z$ is different from $w$. The union $C$ of $S$ and all $S_z$, $z \in D$, is clearly locally connected. Any arc from $w$ in $D$ to the $y$-axis contained in $C$ would have to be contained in $S$ (it intersects each $S_z$ at most in $z$), a contradiction.

Added after reading comments: Let's pick a sequence of planes $\Pi_n$ containing the $y$-axis and converging from above (from the point of view of the half-plane $x\ge 0$) to the $xy$-plane where $S$ resides. Enumerate points $z\in D$ as $z_n$, $n\ge 1$. From each $z_n$ go up until you hit $\Pi_n$ and then create a scaled copy of $S$ lying on $\Pi_n$ and ending on the $y$-axis. That should guarantee $S_z\cap S_w \subset y-\text{axis}$.

Why is $C$ locally connected? The issue is only with point on the $y$-axis. Given any $\epsilon > 0$ and $-1\leq y\leq 1$, consider the union of all $S_z$ that intersect $0\times [y-\epsilon,y+\epsilon]\times 0$ and are of diameter less that $\epsilon$. Add $\epsilon$-neighborhoods of end-points of $S_z$s in $S$. That set is connected, of small diameter, and it contains $(0,y,0)$ in its interior rel.C.

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It's hard for me to see how you keep $S_z$ and $S_w$ from intersecting outside the $y$-axis. If it is convenient to include a picture, that would be a big help... –  Nate Eldredge Nov 22 '10 at 18:56
    
Also, what do you mean by a copy $S_z$ of $S$ in $R^3$? –  Nate Eldredge Nov 22 '10 at 19:00
    
I assume that the idea is to scale and translate S to join z to the y-axis, then bend it a bit in the third dimension to avoid intersections. I don't know why this results in a "clearly locally connected" set though. –  George Lowther Nov 22 '10 at 20:14
    
Jurek, your example almost made me think I understand the problem, but... I don't think $C$ is locally connected at all points of the graph $G\subset S$ of $\sin\frac1x$. Let $p\in G$, and let $p_n$ be a sequence of distinct points in $D$ converging to $p$. Then $S_{p_n}$ Hasdorff-converge to a scaled copy $h(S)$ of $S$ lying in the horizontal plane. Its intersection with $S$ surely contains a $q\in G$ that is far from $p$ and from the $y$ axis. Now $q=h(r)$ for some $r\in G$, and the images $r_n$ of $r$ in $S_{p_n}$ converge to $q$. Any small connected sets containing almost all of the $r_n$? –  Sergey Melikhov Dec 5 '10 at 23:38
    
There is an interesting example here ams.org/journals/bull/1942-48-02/S0002-9904-1942-07615-4 (see Property 7). –  Sergey Melikhov Oct 22 '11 at 20:06
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A partial answer is that this is true for compact sets (as stated in the answer by Mark Sapir) and also for all open sets $M$, because one can prove rather easily, that the set of all points of $M$ which can be joined to some fixed $x_0\in M$ is open and closed at the same time.

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Easier than that: an open set is always locally path connected, because it contains balls centered at every point. –  Nate Eldredge Nov 20 '10 at 17:49
    
Of course. I should think at least three times before I post an answer. I was thinking about nonlocal case, i.e. every open connected subset is path connected. –  Vít Tuček Nov 20 '10 at 19:20
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