Definition of a continuous function

[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?

 

[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..

 

[nicksauce]

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

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

 

[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.

 

[Studiot]

What if x is an isolated point of M?

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

 

[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."

 

[Studiot]

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."

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.

 

[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.

 

[Landau]

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

Therefore this definition says that f is continuous at x0 if every d'''-nhbd of f(x0) contains the image of a d'''-nhbd of x0. Since these nhbd's are open sets, the new definition given in the previous paragraph contains the elementary-calculus definition as a special case.