Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In order to give an example of a function from $\mathbb{R}$ to itself whose graph is connected in $\mathbb{R} \times \mathbb{R}$, yet is not continuous, the book Berkley Problems on Mathematics refers to a function: $f(x)=\sin(\frac{1}{x})$.

Let $$S_1=\left\{(x,\sin\left(\frac{1}{x}\right))\mid x <0\right\}, \ S_2=\left\{(x,\sin\left(\frac{1}{x}\right)) \mid x >0\right\}.$$ $S_1$ and $S_2$ are both connected subsets of $\mathbb{R} \times \mathbb{R}$. Since $(0,0)$ belongs to the closure in $\mathbb{R} \times \mathbb{R}$ of both $S_1$ and $S_2$, the sets $S_1 \cup \{(0,0)\}$ and $S_2 \cup \{(0,0)\}$ are connected. Since these sets have a point in common, their union $$(S_1 \cup \{(0,0)\}) \cup (S_2 \cup \{(0,0)\})$$ is connected. This union is the graph of the function $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x)=\sin(\frac{1}{x})$, $x \neq 0$, $f(0)=0$, which is not continuous at $x=0$.

In this example, what I don't understand is why $(0,0)$ belongs to the closure of $S_1$ and $S_2$. As I know, the function $f(x)$ vibrates strongly near $0$, so how can the graph get near $(0,0)$? How can $(0,0)$ be in the closure of the two sets?

Many thanks!

share|improve this question
add comment

1 Answer 1

For every natural $n$, $S_2$ contains the point $(1/ (n \pi),0)$. As $n$ tends to infinity, these points tend to $(0,0)$, so $(0,0)$ is in the closure of $S_2$. Similarly, $S_1$ contains $(-1/ (n \pi),0$).

share|improve this answer
    
Natural $n$? I think $n$ must be an integer for $(1/(n \pi),0)$ to be in $S_2$... –  shinyasakai Apr 20 '11 at 15:36
    
@shinyasakai: $n$ has to be a natural number; that is, it must be an integer greater than 0. If $n$ is an integer less than zero, then $(1/(n \pi),0)$ is in $S_1$. If $n$ is 0, then $(1/(n \pi),0)$ doesn't make sense at all. –  Chris Eagle Apr 20 '11 at 15:48
    
Oh, I am wrong... When I read and typed "natural", there was "real" in my mind... But, why will the points tend to $(0,0)$ when $n$ tends to infinity? The invariant of the function $f$ can be any real number, not just numbers of the form $1/n \pi$, where $n$ is natural. –  shinyasakai Apr 20 '11 at 15:48
1  
(In a metric space) A point $p$ is in the closure of a set $S$ if there exists a sequence of points $(p_n)\subset S$ such that $p_n \neq p$ and $|p_n -p| \to 0$ as $n\to \infty$. You seem to be under the impression that the convergence condition needs to hold for all sequences, an impression that is false. –  Willie Wong Apr 20 '11 at 17:01
2  
Alternatively, any neighborhood of (0,0) intersects both $S_1$ and $S_2$, making it a limit point of each of the sets, so that (0,0) must belong to the closure of both. –  gary Apr 20 '11 at 18:28
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.