Definition of a continuous function

1. Jun 12, 2010

nicksauce

I am reading Schutz's "Geometrical methods of mathematical physics". He writes: "A map f:M->N is continuous at x in M if any open set of N containing f(x) contains the image of an open set of M." However, it seems to me that a more appropriate definition would be "... contains the image of a neighbourhood of x". Am I right, or am I missing something obvious?

2. Jun 12, 2010

arildno

I would think that with Schutz's definition, you'll be able to PROVE that neighbourhoods around "x" must be included in those "open sets of M".

Note, for example, that a discontinuity of value at f(x) will prevent the existence of there being any open set of N about it. I think..

3. Jun 12, 2010

nicksauce

Well the counter-example of the Schutz definition I'm thinking of would look something like this: http://imgur.com/ppE5t [Broken]

But maybe I need to think about it some more...

Last edited by a moderator: May 4, 2017
4. Jun 12, 2010

Tedjn

Counterexample perhaps from R to R:
$$f(x) = \begin{cases} 1 & \text{if }x \in \mathbb{Q}\text{ and }x \geq 0, \\ 0 & \text{if }x \notin \mathbb{Q}\text{ and }x \geq 0, \\ 1 & \text{if }-1 < x < 0, \\ 0 & \text{if }x \leq -1. \end{cases}$$​
According to the given definition, f(x) is continuous everywhere.

5. Jun 12, 2010

Studiot

What if x is an isolated point of M?

Was the definition not so written to cater for this eventuality?

6. Jun 13, 2010

snipez90

No, that's not the issue. Clearly if you consider continuity at x and then forget to require an open set about x to be contained in an arbitrarily small open set about f(x), you have a worthless definition. I'm pretty sure the author meant to say "A map f:M->N is continuous at x in M if any open set of N containing f(x) contains the image of an open set of x in M."

7. Jun 13, 2010

Studiot

That would satisfy nicksauce' qualms, without introducing a neighbourhood.

However the original question was should a neighbourhood be introduced and I think I answered that point since I don't think an isolated point has a neighbourhood, although it is defined to be an open set, thereby satisfying your form of words.

I found, like others, it quite hard to get my head round this particular statement of continuity, which differs subtly from the topological one I am more used to.

So I think you may have cracked it.

8. Jun 13, 2010

jostpuur

This is a simple counter example.

$$f:\mathbb{R}\to\mathbb{R}$$

$$f(x) = x,\quad\quad x\neq 0$$

$$f(0) = 1$$

With Euclidian topology, $f$ is not continuous at $x=0$. Now choose an arbitrary open set $V\subset\mathbb{R}$ such that $1\in V$ (that means $f(0)\in V$). You can find an open set $]1-\epsilon,1+\epsilon[$, such that $f(]1-\epsilon,1+\epsilon[)\subset V$, so $f$ should be continuous at $x=0$ according to the strange definition now.

9. Jun 13, 2010

Landau

It is a mistake. But Schutz says it right at the following page (8):