Definition of discontinuous function

bogdanno · P versus NP Offline Joined: 23 Jun 2008 Posts: 46

I have seen in a book or here http://www.toolingu.com/definition-800310-41529-discontinuous-function.html the following case of discontinuity.
If a function is not defined at a point, then it is discontinuous at that point. Is that true? In my mind, if the definition of continuity is applied on the domain of the function, then discontinuity is also applied on the domain. Why would you make affirmations about points which are not in domain?

**Posted:** Tue Oct 06, 2009 11:43 pm

-Elixir- · Poincare Conjecture Offline Joined: 15 May 2009 Posts: 104

A function $f$ is continuous at $a$ if and only if $\lim_{x \to a}f(x) = f(a)$ . This implies three things:

$1) f(a)$ exists
$2) \lim_{x \to a} f(x)$ also exists
$3) \lim_{x \to a} f(x) = f(a)$

Since your $f$ fails to satisfy $1)$ at $a$ , $f$ is discontinuous at $a$ .

Now, to answer your question, this is my take: When you say that a function is discontinuous at $a$ , you are not talking about a point on the graph (which, in this case, does not exist) but the line $x=a$ ; the function breaks as it crosses that line. Does this make things clearer?

**Posted:** Wed Oct 07, 2009 1:13 am

bogdanno · P versus NP Offline Joined: 23 Jun 2008 Posts: 46

Yes, it is clear. But I think the problems that ask to determine the discontinuities where the function is not defined are bad problems. The notion of continuity is essentially more than that. It might confuse the student who just learned about the limits and continuous functions.
For example, consider this problem:
"Determine the point(s) at which the given function is not continuous.
f(x)= 5 csc (7x) "
Why wouldn't we ask the same for differentiable functions?

**Posted:** Thu Oct 08, 2009 6:34 pm

t0rajir0u

The "AP calculus" definition says that a function is discontinuous at a point if it isn't defined there. But I agree that a student who wants to take real analysis and/or topology seriously should only consider continuity with respect to the subspace topology on the domain on which the function is defined.

**Posted:** Thu Oct 08, 2009 9:03 pm

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hsbhatt

This has been a source of confusion to me too, because for f to be called a function, it must be defined at every point in its domain. So what is the point in discussing discontinuity at a point not in the domain at all

**Posted:** Tue Oct 13, 2009 8:03 pm

JBL

Nearly every function you're interested in when studying calculus can be defined over all of $\bf R$ . It seems to me useful to have a way of distinguishing those functions that aren't so-defined.

**Posted:** Wed Oct 14, 2009 4:02 am

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mavropnevma

I think it is ridiculous for a textbook to discuss, define, consider, etc, discontinuity at a point not in the domain of the function. The only issue that may occur (and it is an important issue) is if the function may be prolonged by continuity to a point not in its domain.

For example, $f : \mathbb{R} \setminus \{0\} \to \mathbb{R}$ given by $f(x) = x\sin\frac {1} {x}$ may be prolonged to a continuous function $\tilde{f}$ on $\mathbb{R}$ by taking $\tilde{f}(0) = 0$ , while $g : \mathbb{R} \setminus \{0\} \to \mathbb{R}$ given by $g(x) = \sin\frac {1} {x}$ may not.

**Posted:** Wed Oct 14, 2009 5:43 am

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b555 · Poincare Conjecture Offline Joined: 07 Mar 2009 Posts: 221

pardon me but

The term removable discontinuity is sometimes (improperly) used for cases in which the limits in both directions exist and are equal, while the function is undefined at the point x0

is it not true??

**Posted:** Wed Oct 14, 2009 6:52 am

t0rajir0u

Yes, but that's not the issue. The issue is when you claim that, for example, $\frac{1}{x}$ is discontinuous at $x = 0$ .

**Posted:** Wed Oct 14, 2009 8:28 am

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fedja

**Posted:** Fri Oct 16, 2009 5:28 pm

mlok

**Posted:** Wed Oct 28, 2009 8:20 am

fedja

But if you restate it as "For every sequence $x_k$ in the domain such that $x_k\to a$ , one has $\lim_{k\to\infty}f(x_k)=f(a)$ ", you'll be fine again, even at the isolated points. Smile

**Posted:** Wed Oct 28, 2009 1:51 pm

gauss202

In this definition, fedja, would you assume that $a$ is in the domain of $f$ , or allow it to possibly not be?

**Posted:** Wed Oct 28, 2009 1:59 pm

mlok

If $a$ wasn't in the domain, what would $f(a)$ even mean?

**Posted:** Wed Oct 28, 2009 2:14 pm

pankajsinha

It is nice that this discussion has taken place.I had posted this question (refer to http://www.artofproblemsolving.com/Forum/viewtopic.php?t=259134) and I had this very doubt in my mind,but the discussion did not materialise.

**Posted:** Thu Oct 29, 2009 4:06 am

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**Posted:** Mon Nov 02, 2009 2:21 am

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