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Is the set $\{\big(x,\sin(1/x)\big):x\ne 0 \}$ connected in usual metric of $\mathbb R^2$ ?

I tried writing it as a union of two connected sets , or otherwise as a union of two disjoint non-empty open sets without any success , Please help . Thanks in advance

Mikasa
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  • It is not connected. A disconnection is made up by considering the two components one in the left open half-plane, the other on the right one. – Crostul Mar 17 '16 at 15:45
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    ${\left(x,\sin(1/x)\right):x>0 }$ and ${\left(x,\sin(1/x)\right):x<0 }$ seem open to me... – Arnaud D. Mar 17 '16 at 15:45
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    http://math.stackexchange.com/q/317125/186473 – Learnmore Mar 17 '16 at 15:46
  • @ArnaudD. : How are they open ? –  Mar 17 '16 at 15:48
  • @learnmore : that doesn't help .. –  Mar 17 '16 at 15:51
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    @SaunDev Recall that, given a subset $A$ of $\Bbb R^n$, a subset $B\subseteq A$ is open in $A$ if it is an intersection of $A$ and a set open in $\Bbb R^n$. – Wojowu Mar 17 '16 at 15:54
  • @learnmore : the union of two parts $A$ and $B$ is not connected –  Mar 17 '16 at 15:55
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    @SaunDev They are intersections of your set with $(0,+\infty)\times \mathbb{R}$ and $(-\infty,0)\times\mathbb{R}$ respectively, and these are open in $\mathbb{R}^2$. Thus they are open subsets of your set for the induced topology (but not open sets in \mathbb{R}^2$). – Arnaud D. Mar 17 '16 at 16:20

3 Answers3

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The image of $A$ under the continuous projection $(x, y) \to x$ is $\mathbb{R}\setminus\{0\}$, which is not connected. Hence $A$ itself is not connected.

anomaly
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Recall that a connected and locally path-connected topological space is path-connected. Your subspace $X=\left\{\left(x,\sin\frac1x\right)\,:\,x\ne0\right\}$ is locally path-connected. However, you can easily convince yourself (in fact, if you asked the question, you probably already are) that no curve with image contained in $X$ joins $\left(\frac1\pi,0\right)$ with $\left(-\frac1\pi,0\right)$.

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Let $A=\{\big(x,\sin(1/x)\big):x\ne 0 \}$. Let us define $f: R-{0} \rightarrow R^2$ such that $f(x)=(x,\sin(1/x))$, then $f$ is continuous on $R-{0}$. Now $A=f(R^+) \bigcup f(R^-)$(where $R^+$ is the +ve reals and $R^-$ is the -ve reals). Now each of $f(R^+)$ and $f(R^-)$ are disjoint and connected (as both are continuous image of a connected set). So they are two components of $A$, hence $A$ is not connected.

Sourjya Banerjee
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  • You also need to show that each of $f(\mathbb R^+)$ and $f(\mathbb R^-)$ is open (or closed). – Ayman Hourieh Mar 17 '16 at 18:37
  • Here I am using the fact that a topological space is connected iff it has only one component. Where a component of topological space X is a connected subset of X which is not properly contained in a connected subset of X. – Sourjya Banerjee Mar 17 '16 at 18:51
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    Well, $\mathbb R = (-\infty, 0) \cup [0, \infty)$. These intervals are connected and disjoint. You need more in order to show that each subset is a distinct component. – Ayman Hourieh Mar 18 '16 at 01:49