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I posted a solution here with an illustration (see below) and commented that the function was discontinuous at $x=1$, where it is undefined. Someone told me, no, it is undefined but continuous.

Now I'm confused.

I would have thought that the point $(1, -1/2)$ in the graph below should be designated with a hollow point (a point that isn't filled in) to demonstrate that that point is not actually on the graph. Furthermore, I thought that this hole would constitute a discontinuity.

If I need to be set straight here, could someone please help me out?

undefined point

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    $\begingroup$ For sure,it should be marked with a circle because the point x=1 is not in the domain. $\endgroup$ – Suraj_Singh Sep 12 '15 at 6:07
  • $\begingroup$ I added the clarification that someone had told me that it is undefined at this point, but continuous. Wouldn't the circle be a discontinuity? $\endgroup$ – Adam Hrankowski Sep 12 '15 at 6:09
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    $\begingroup$ It has to be a discontinuity.If it's not (suppose) then for continuity we must have value of function at a point= limit at that point. So in the present case, limit exists but value doesn't so function is discontinuous at x=1. $\endgroup$ – Suraj_Singh Sep 12 '15 at 6:14
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    $\begingroup$ What the person could have meant was that the function itself is continuous which simply means that it's continuous in all points where it's defined $\endgroup$ – Alice Ryhl Sep 12 '15 at 6:44
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    $\begingroup$ Your function is not defined at $x=1$. Thus it does not make sense to say that $f$ is continuous (or discontinuous) at $x=1$. $\endgroup$ – user99914 Sep 12 '15 at 6:47
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A function from a set $A$ to a set $B$ is a certain subset of their Cartesian product. Any property of a function must therefore refer to its defining data. As such, continuity or lack thereof can only be determined in the domain of the function.

This is NOT pedantry. The function $f$ defined on $\mathbb{R}\setminus \{0\}$ which takes the value $1$ for $x>0$ and $-1$ for $x<0$ is a continuous function. There is no "in its domain" qualifier. It is a continuous function, period. In the OP's example, the function is perfectly continuous and even has a continuous extension to the entire real line.

Edit: now that the other answer is edited, the clarification is no longer necessary.

Edit 2: Justify downvotes.

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  • $\begingroup$ Your argument reminds me of the point made here about limits. The post makes a similar point about restricting the test to the domain of the function.math.stackexchange.com/a/1429451/266049 $\endgroup$ – Adam Hrankowski Sep 12 '15 at 14:09
  • $\begingroup$ I found your aside about the extension helpful as well. $\endgroup$ – Adam Hrankowski Sep 12 '15 at 14:11
  • $\begingroup$ Thanks much. I love this stuff. We're such geeks. ;) $\endgroup$ – Adam Hrankowski Sep 12 '15 at 15:53
  • $\begingroup$ It sounds like you are saying we can speak of a function's being discontinuous at a point outside its domain. Wouldn't that be trivial? What I find more interesting is where a function is discontinuous within its domain. $\endgroup$ – Adam Hrankowski Sep 12 '15 at 16:11
  • $\begingroup$ @AdamHrankowski Yes, your textbook definition allows to say a function is not continuous (I don't want to say is discontinuous, it looks like a positive statement and adds to confusion!) at a point $c$ outside its domain. Of course generally we only care about what happens in the domain, but there are cases, like yours, where the function can be extended outside of its domain, and then these extensions become interesting. In ODE for instance, we can sometimes push a solution outside the domain it is originally defined, so those points we found uninteresting before become interesting now :) $\endgroup$ – guest Sep 12 '15 at 16:19
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If a function is undefined at a point, then you can't speak of it being either continuous or discontinuous there. Those terms are only defined for points in the domain of the function. Stein and Barcellos, Calculus and Analytic Geometry, 5th Edition (Sec. 2.8):

According to this definition any polynomial is continuous. So is each of the basic trigonometric functions, including $y = \tan x$... You may be tempted to say, 'But $\tan x$ blows up at $π/2$ and I have to lift my pencil off the paper to draw the graph.' However, $π/2$ is not in the domain of the tangent function... If $a$ is not in the domain of $f$, we do not define either continuity or discontinuity there.

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Excerpt from my answer to this question:

The term continuous function is defined with respect to its domain. Therefore it is crucial to specify the domain of a function, if we want to analyse the function with respect to continuity. Outside of the domain of a function this function is not continuous, since it's not even defined there.

Note that when we talk about discontinuities of a one variable function we classify them as either being a removable discontinuity, a jump discontinuity or an essential resp. infinite discontinuity. The key point here is, that each of these discontinuities is defined with respect to the domain of $f$. We conclude, discontinuities are defined solely within the domain of $f$.

Informally: The domain and codomain specify where the function lives and we can't say anything about the function outside of its region of existence.

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