This discussion board is a place to discuss our Daily Challenges and the math and science
related to those challenges. Explanations are more than just a solution — they should
explain the steps and thinking strategies that you used to obtain the solution. Comments
should further the discussion of math and science.
When posting on Brilliant:
Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.
Markdown
Appears as
*italics* or _italics_
italics
**bold** or __bold__
bold
- bulleted - list
bulleted
list
1. numbered 2. list
numbered
list
Note: you must add a full line of space before and after lists for them to show up correctly
I found this discussion on internet. May be this is where you should discuss it. :)
http://math.stackexchange.com/questions/455296/can-a-function-with-just-one-point-in-its-domain-be-continuous
Hey Ivan, is the given function a function at all ? I'm asking bcoz the ordered pairs ( i.e the function values in the cartesian plane) is of the form (0,R) R=real no. . All the ordered pairs have the same 1st element that is 0. So am I confusing something here?
@Kulkul Chatterjee
–
No no.. he is talking about case where you pick a particular real number. So you will have only one pair (0,r) and then indeed it is a function.
Indeed. If the domain contains a single element, then the inverse image f−1(U) of any set is either empty or the whole domain. Both of these sets are open in the domain. Thus the inverse image of every set is open, and hence the inverse image of every open set is open, which means that f is continuous. This result holds no matter what set the codomain is, and no matter what topology it is equipped with.
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*
or_italics_
**bold**
or__bold__
paragraph 1
paragraph 2
[example link](https://brilliant.org)
> This is a quote
\(
...\)
or\[
...\]
to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
Reframe and elaborate your question.
Log in to reply
The question is whether a function with a single-element domain is continuous. For example, is any function f:{0}→R continuous?
In my opinion, it's vacuously true, but I'm not sure.
Log in to reply
I found this discussion on internet. May be this is where you should discuss it. :) http://math.stackexchange.com/questions/455296/can-a-function-with-just-one-point-in-its-domain-be-continuous
Hey Ivan, is the given function a function at all ? I'm asking bcoz the ordered pairs ( i.e the function values in the cartesian plane) is of the form (0,R) R=real no. . All the ordered pairs have the same 1st element that is 0. So am I confusing something here?
Log in to reply
(0,r) and then indeed it is a function.
No no.. he is talking about case where you pick a particular real number. So you will have only one pairIndeed. If the domain contains a single element, then the inverse image f−1(U) of any set is either empty or the whole domain. Both of these sets are open in the domain. Thus the inverse image of every set is open, and hence the inverse image of every open set is open, which means that f is continuous. This result holds no matter what set the codomain is, and no matter what topology it is equipped with.