Let $A = \{ (x,y) \in \Bbb{R}^{2} \mid y = mx \} \setminus \{ ( 0,0 ) \} \subset \Bbb{R}^{2}$. Then $A$ is

(a) open,

(b) closed,

(c) connected,

(d) nowhere dense.

This question was asked in NBHM test for PhD scholarship 2005 (I think they forgot to mentioned that $m$ is a real number).

The correct answer is (d) nowhere dense (according to the answer key they have provided).

For (b): Here I can prove that given set is not closed since $(0,0)$ is a limit point of this set which doesn't belong to the set.

For (a): Intuitively I'm feeling that it is not open, since this set is set of lines passing through the origin in $\Bbb{R}^{2}$. We are unable to draw open ball at any point of the set which is contained in the set. (Is my reasoning correct? Please mention if there is another method to prove that the set is not open).

For (c): This set is the union of disjoint sets, since the intersection of lines passing through origin is singleton set $\{(0,0)\}$ but here that point doesn't belong to the set. but I don't think this argument is enough to prove given set is not connected. Since we know the example of broken comb space which can be written as union of disjoint sets but broken comb space is connected. (I know that there are several other methods of proving given set not connected for eg: there will be a continuous function from given set to discrete space $\{0,1\}$, but I'm unable to construct such function).

For (d) Given set is union of straight lines passing through origin (line in real plane is nowhere dense) and since real plane is complete under usual metric, by Baire's theorem given set is nowhere dense. Is my argument correct?

Please help.

Thanks in advance.

  • 3
    $\begingroup$ Sorry if the question is so useless but, is $m$ a fixed real number? $\endgroup$ Jul 17 '17 at 8:03
  • 2
    $\begingroup$ @CarlosJiménez OMG ! I never thought of this ! actually it can be a fixed real number , and then the answer will be so obvious ! But since they haven't mentioned it I got confused. Because we use y=mx for any line through the origin. $\endgroup$ Jul 17 '17 at 8:24
  • 2
    $\begingroup$ I think it's just a single line. $\endgroup$ Jul 17 '17 at 8:25
  • 1
    $\begingroup$ @HennoBrandsma yes ! big facepalm moment for me! $\endgroup$ Jul 17 '17 at 8:27
  • 2
    $\begingroup$ As mentioned above the set in question is a single line. I'm not deleting the question since it will always remind me "read the question carefully before wasting energy and time on it ". $\endgroup$ Jul 17 '17 at 8:45

I think $m$ is fixed.

Any line in the plane is connected, closed and nowhere dense and it contains no non-empty open set of the plane. Leaving out the origin kill connectedness (like the reals minus a point) and closedness (as the origin is still a limit point, but no longer in the set). Subsets of nowhere dense sets are still nowhere dense, so indeed (d) must be correct.

If we take the union over all $m$ then we get $\mathbb{R}^2$ minus the $y$-axis which is open, not closed, not connected and not nowhere dense.. This seems unlikely to me.

  • $\begingroup$ Thank you! I thought this set is union of all lines through origin :p $\endgroup$ Jul 17 '17 at 8:30
  • 1
    $\begingroup$ @SushantPawar you always miss the $y$-axis though, a it is not of the form $y = mx$ (but $x=0$). $\endgroup$ Jul 17 '17 at 8:31
  • $\begingroup$ yes actually! while thinking about the question ( wrong question) somewhere I felt the same about the y axis. Thank you so much! :) +1 for this! $\endgroup$ Jul 17 '17 at 8:34
  • 1
    $\begingroup$ @SushantPawar glad I could help... $\endgroup$ Jul 17 '17 at 8:35
  • $\begingroup$ it taught me a new lesson of reading the question carefully before starting thinking about it and wasting energy on it! :D $\endgroup$ Jul 17 '17 at 8:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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