## Excel in math and science

### Master concepts by solving fun, challenging problems.

## It's hard to learn from lectures and videos

### Learn more effectively through short, conceptual quizzes.

## Our wiki is made for math and science

###
Master advanced concepts through explanations,

examples, and problems from the community.

## Used and loved by over 5 million people

###
Learn from a vibrant community of students and enthusiasts,

including olympiad champions, researchers, and professionals.

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) 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

Sort by:

TopNewestReframe 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 \(f : \{0\} \to \mathbb{R}\) continuous?

In my opinion, it's vacuously true, but I'm not sure.

Log in to reply

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.

Log in to reply

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

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

Log in to reply