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Find the connected components of the following sets: $(a) \; A=\{(x,y):y=\sin(1/x), x\in\mathbb{R}^+\},(b)\;A\cup\{(x,y):x=0,y\in[-1,1]\},(c)\;$The Cantor Set,$(d)\;\mathbb{N}$ with the cofinite topology.

I tried to solve them as follows, I'm unsure if what I did is correct

$(a)$ $\sin(1/x)$ is continuous in $(0,+\infty)$ and $\mathbb{R}^+$ is connected. Since any continuous image of a connected space is connected, the set $A$ is connected. Being $A$ connected, the set has only one component, the set $A$ itself.

$(b)$ The vertical segment is connected but has no intersection with the set $A$, then their union cannot be connected but each set ($A$ and the vertical segment) are the connected components of the space.

$(c)$ The Cantor Set is totally disconnected and since the only connected spaces in the real line are intervals, the components of the space are the ones of the form $\{x\}$.

$(d)$ I have no clue. If I had to give an answer, I'd is connected because the open sets here have the form $\{\dots,m-2,m-2,m\}\cup\{n,n+1,n+2\}$ then the sets of the form $\{a,a+1,\dots,a+k\}$ are closed sets and $\mathbb{N}$ can't be written as the union of two closed set of the last kind (taking $\{a,\dots, a+k_1\}\cup\{b,\dots,b+k_2\}$ $(a-1)$ is not in the set).

What further argument could a give to $(b)$?, and are the answers ok? (specially (d), which left me baffled).

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(b) is called the topologist's sine curve and is actually connected, but not path connected. This is since you need to separate your space into two disjoint open sets to show it is not connected. –  rVitale May 22 at 21:03
    
@rVitale Are the same curve?. I gave it a look to the topologist's sine curve but the definition I found only includes the point $(0,0)$ instead of $\{(x,y):x=0,-1\leq y \leq 1\}$. I guess that this makes no difference and is still connected but not path connected?. So being connected it has only one component? –  Cure May 23 at 1:19
    
there are some variations to the topologist's sine curve and both the one in your post and the one you mentioned above are among the variations. Yes, being connected it has only one component. Since connected components are maximal connected subsets, there can be no smaller connected component as it would be a subset of the connected space. –  rVitale May 23 at 11:29

1 Answer 1

up vote 1 down vote accepted

In the case of (d) $\mathbb{N}$ is connected, since suppose we had a separation into two nonempty open disjoint subsets: $\mathbb{N}=U \cup V$. We know $U= \mathbb{N}-\{n_1,...,n_k\}$ and that $V=\mathbb{N}-\{m_1,...,m_l\}$. Now $U$ and $V$ can't be disjoint, as we can pick some natural number $n \notin \{n_1,...,n_k,m_1,...m_l\}$ and it is in both $U$ and $V$.

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