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Why a connected subspace of a locally connected space X is locally connected if X is the real line?

Is this true if X is an arbitrary locally connected space?

Thanks for your help

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    What have you tried so far, Daniela? Remember, explaining what your thoughts are about the problem will help to prevent people from telling you things you already know, help them write their answers at an appropriate level, and moreover will make people much more willing to help.2012-06-14

2 Answers 2

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This is true if $X=\mathbb{R}$ because connected subsets in this case are precisely the intevals.

For a counterexample for other spaces take $X=\mathbb{R}^2$ and $Y=\{ (x,y): x=0,\ 0\leq y\leq 1\} \cup \{ (x,y) :x\in [0,1],\ y=\sin\left(\frac{1}{x}\right) \}=Y_1\cup Y_2$. Then points in $Y_1$ don't have connected open neighbourhoods (in $Y$).

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HINTS:

  1. Every non-empty connected subset of $\Bbb R$ is homeomorphic to one of four subsets of $\Bbb R$. Two of these are $\{0\}$ and $\Bbb R$ itself. The other two are also very simple and familiar; what are they? Once you’ve found them, you should be able to see easily that all four are locally connected, and therefore every subspace of $\Bbb R$ is locally connected.

  2. Let $X=\Bbb R^2$ and $S=\{0\}\cup\{1/n:n\in\Bbb Z^+\}$. Let $Y=\Big([0,1]\times\{0\}\Big)\cup\Big(S\times[0,1]\Big)\;.$ Show that $Y$ is connected. Is $Y$ locally connected? (Consider neighborhoods of the point $\langle 0,1\rangle$.)

The space in (2) is sometimes known as the comb space; it’s a useful source of examples involving connectedness-type properties.