0

"We defined the notion of continuity for functions defined on a subset $E$ of a metric space $X$. However, the complement of $E$ in $X$ plays no role whatever in this definition (note that the situation was somewhat different for limits of functions). Accordingly, we lose nothing of interest by discarding the complement of the domain of $f$. This means that we may juct as well talk only about continuous mappings of one matric space into another, rather than of mappings of subsets."

I have one question:

Why $E^c$ in $X$ plays no role in definition of continuous function but it's material in definition limit of function?

RFZ
  • 16,814

1 Answers1

1

Consider the function $f:\Bbb R\setminus\{0\}\to\Bbb R$ given by $$f(x)=\frac{x^2}{x}.$$ Clearly, $f$ isn't defined at $0,$ and is continuous. Moreover, $\lim_{x\to0}f(x)$ exists.

By contrast, $g(x)=\frac{|x|}x$ has the same domain and is continuous, but $\lim_{x\to0}g(x)$ does not exist.

So, clearly, we can talk about limits at points outside of a function's domain, sometimes. However, when we talk about continuity, we cannot do this. A function can only be said to be (dis)continuous at points of its domain, sonce it doesn't exist outside of its domain, so can't be said to be anything at all there.

Added: The issue ultimately comes down to the metric we're using for the domain. First, let me address this in the context of a specific example I gave above, namely: $X=Y=\Bbb R,E=\Bbb R\setminus\{0\},$ $f$ defined as above, $d_X:X\times X\to\Bbb R$ given by $d_X(x,y)=|x-y|,$ $d_Y$ defined similarly, and $d_E$ the restriction of $d_X$ to $E\times E.$

Now, we can say that $f:E\to Y$ is continuous on $E$ with respect to the metric $d_X,$ meaning that $$\forall\epsilon>0,\forall c\in E,\exists\delta>0:\forall x\in E,d_X(x,c)<\delta\implies d_Y\bigl(f(x),f(c)\bigr)<\epsilon.$$ We can also say that $f$ is continuous on $E$ with respect to $d_E,$ simply by replacing "$d_X(x,c)<\delta$" with "$d_E(x,c)<\delta$" in the statement above. This should come as no surprise--after all,, for any $c,x\in E,$ we have $d_E(x,c)=d_X(x,c)$ by definition of $d_E.$ In that sense, then, it makes no difference whether we consider the domain of $f$ as the entire space, or simply a subset of some larger space. When talking about limits, however, it is a different story.

Clearly, for any $c\in X,$ we can say that $\lim_{x\to c}f(x)=c$ with respect to $d_X,$ meaning $$\forall\epsilon>0,\exists\delta>0:\forall x\in E,0<d_X(x,c)<\delta\implies d_Y\bigl(f(x),c\bigr)<\epsilon.$$ Also, for any $c\in E$ (Note: this is the part that's different), we can say that $\lim_{x\to c}f(x)=c$ with respect to $d_E,$ by replacing "$0<d_X(x,c)<\delta$" with "$0<d_E(x,c)<\delta$" in the statement above. It seems like a superficial change, so why can't we say it for all $c\in X$? Well, this is simply because if $c\in X\setminus E$ (that is, if $c=0$), then $d_E(x,c)$ is undefined for all $x.$

More generally, suppose $\langle X,d_x\rangle,\langle Y,d_Y\rangle$ are any metric spaces, $E\subseteq X,$ $d_E$ the restriction of $d_X$ to $E\times E,$ and $f:E\to Y.$ Then $f$ is continuous at a point $c\in E$ with respect to $d_X$ if and only if it is continuous at $c$ with respect to $d_E.$ So, continuity doesn't change if we start thinking in terms of $\langle E,d_E\rangle$ instead of $E$ being a part of a larger space. However, given $c\in X\setminus E,$ while we may be able to talk about $\lim_{x\to c}f(x)$ with respect to $d_X,$ we will certainly not be able to talk about $\lim_{x\to c}f(x)$ with respect to $d_E.$ So, if we start thinking in terms of $\langle E,d_E\rangle,$ then we may lose some limits of $f$ that we had when thinking of $E$ as part of a larger space.

Cameron Buie
  • 102,994
  • But what about function $h(x)=\text{sgn}x$ in $E=[1,2]\subset \mathbb{R}$. We see that $f(x)$ is continuous at $E$ but in $E^c$ it's not continuous. – RFZ Sep 09 '15 at 19:13
  • Very true, but note that the test says defined on a subset $E.$ By this, they mean that $E$ is the domain of the function. But $[1,2]$ is not the domain of $\operatorname{sgn} x.$ It is the domain of the restriction of $\operatorname{sgn} x$ to $[1,2],$ of course, but that restriction is not defined outside of $[1,2].$ Let me adjust my answer to elucidate. – Cameron Buie Sep 09 '15 at 19:37
  • See also the related questions here and here. – Cameron Buie Sep 09 '15 at 20:01
  • Why that restiction is not defined outside of $[1,2]$? – RFZ Sep 10 '15 at 04:40
  • Because we restricted it. By definition of restriction, it is only defined on $[1,2].$ – Cameron Buie Sep 10 '15 at 04:46
  • Am I true that remrk says that function will be continuous at metric space $(E,d_E)$? – RFZ Sep 10 '15 at 04:52
  • I'm not sure what you're asking. (Which remark? Which function?) Can you clarify for me? – Cameron Buie Sep 10 '15 at 04:56
  • Dear Cameron Buie! I read your post again. My opinion is such: Let $E=\text{Dom}(f)$ then we CAN'T talk about continuous of $f$ outside of $E$. But it's possible for limits fucntion (For example, $h(x):\mathbb{R^1}-{0}\to {1}$ we see that $h(0)$ is undefined but limit in $0$ exists). Am I true? – RFZ Sep 10 '15 at 09:21
  • 1
    I'm still not quite sure I understand you correctly, but I think you have the right idea. Regardless, I've expanded my answer to (hopefully) make it clearer. Let me know what you think! – Cameron Buie Sep 10 '15 at 18:02
  • I read your edited post. It's writen amazing and I understood all moments that I wanted. Thank you very much dear Cameron! It's great! – RFZ Sep 10 '15 at 18:40
  • You are very welcome! – Cameron Buie Sep 10 '15 at 19:36
  • This is very late, but I am currently reading this same remark by Rudin and am confused just like the OP was. I therefore would like to confirm -- is the distinction simply that limits are defined for limit points of the subset whereas continuity is only defined for points of the subset? – EE18 Jun 14 '21 at 12:20
  • @1729_SR: Yes. Continuity and discontunuity can only occur at points of a function's domain (and discontinuity can only occur at non-isolated points of the domain), but limits can exist at points outside of the domain (so long as they are limit points of the domain). – Cameron Buie Jun 14 '21 at 12:32
  • Fair play, thanks very much Cameron. I guess the confusion is heightened by the fact that limits are not necessarily defined at points in the domain (isolated points in particular), so the notion of continuity does not sit perfectly inside that of limits (at least as defined by Rudin). – EE18 Jun 14 '21 at 12:39
  • 1
    @1729_SR: That's fair to say. The notions of continuity and limits have quite a bit of overlap, but neither encompasses the other. Arguably, discontinuity has more to do with the notion of limits than continuity does! – Cameron Buie Jun 14 '21 at 12:45