Suppose $c$ is an isolated point in the domain $D$ of a function $f$. In the delta neighbourhood of $c$, does the function $f$ have the value $f(c)$?

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    $\begingroup$ If delta is small, which elements of D are going to be in a delta neighborhood of c? $\endgroup$ – Jonas Meyer Dec 27 '10 at 5:48
  • $\begingroup$ @Jonas:If no elements are there in the delta neighbourhood of c, then f is not defined at this delta heighbourhood no. $\endgroup$ – Vinod Dec 27 '10 at 5:54
  • $\begingroup$ It never happens that the delta neighborhood is empty. After all, it contains c. $\endgroup$ – Jonas Meyer Dec 27 '10 at 5:55
  • $\begingroup$ @Jonas:Understood. It contains c ONLY. $\endgroup$ – Vinod Dec 27 '10 at 5:58
  • $\begingroup$ Yes, for sufficiently small delta. Then you can see all values f can have in such a neighborhood. $\endgroup$ – Jonas Meyer Dec 27 '10 at 5:59

You can also see this is true using the topological definition of continuity at a point: a function is continuous at a point $f(x)$ if for any neighborhood $V$ of $f(x)$ there is a neighborhood $U$ of $x$ such that $f(U)$ is contained in $V$. For an isolated point, you can take the neighborhood consisting of just the point $c$, so its image $f(c)$ will obviously be contained in $V$, as $V$ is a neighborhood of $f(c)$.

  • $\begingroup$ Please enlighten me: I fail to see how $f: \mathbb{N} \mapsto \mathbb{R} | f(n) = \frac{1}{n}$ is continuous at every point in $\mathbb{N}$. (Let's say $0\not\in\mathbb{N}$.) $\endgroup$ – chharvey Apr 3 '12 at 1:27
  • $\begingroup$ $f$ is continuous if the preimage of any open set in $\mathbb{R}$ is open. Assuming you give $\mathbb{N}$ the topology inherited from $\mathbb{R}$, every set in $\mathbb{N}$ is open, so the statement is vacuously true. To show that any set is open, consider the set $(n-1/2, n+1/2)\cap\mathbb{N}$ for any $n\in\mathbb{N}$. This is an open set; it's also equal to $\{n\}$. So, $\{n\}$ is open for all $n$; the fact that every set is open follows directly (in other words, $\mathbb{N}$ inherits the discrete topology from $\mathbb{R}$). $\endgroup$ – Ben Lerner Apr 4 '12 at 3:33
  • $\begingroup$ I'm not sure what all this has to do with continuity at isolated points. Here's a counterexample: Let $A=[0,1]\cup\{2\}$ and $f(x)=\big\{0:x\in[0,1];1:\text{otherwise}\big\}$. Choose $c=2$ and $\epsilon=\frac{1}{2}$. No matter which $\delta$ is chosen, $0<|x-c|<\delta$ implies $|f(x)-f(c)|=1\not<\epsilon$. Therefore $f$ is not continuous at $c$, which is an isolated point in $A$. $\endgroup$ – chharvey Apr 14 '12 at 19:51
  • $\begingroup$ That counterexample is wrong - let's examine it closely. Take $f:A\to\mathbb{R}$ and see if it's continuous at 2. Take any open set containing 1 in $\mathbb{R}$, and take the open set with $\delta = 1/2$ around 2 in $A$. This is just $\{2\}$, so $f(2) = 1$ is contained in the open set in $\mathbb{R}$. So, $f$ is continuous at 2. (Obviously, if you take $f:\mathbb{R}\to\mathbb{R}, it's not continuous at 2.) $\endgroup$ – Ben Lerner Apr 16 '12 at 3:00
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    $\begingroup$ Fuck me, never mind. I'm mistaking the definition of continuity with the definition of a limit. You're right, you are allowed to take $x=c$; and $|x-2|$ need not be greater than 0. So if $|x-2|<\delta$, and we choose any $0<\delta<1$, then $x\in\{2\}$, aka $x=2$, thus $|f(x)-f(c)|=0<\epsilon$ and $f$ is continuous at 2. :) $\endgroup$ – chharvey Apr 16 '12 at 6:15

I see now that the comments above provide essentially the answer, with whatever definition of continuity you have. The following ties everything together.

Let's use as the definition of continuity $\lim_{x \rightarrow c}\,f(x) = f(c)$. Expand: For all $\varepsilon > 0$ there exists $\delta > 0$ such that whenever $0 < |x-c| < \delta$ it is true that $|f(x)-f(c)| < \varepsilon$. When $\delta$ is small enough, there are no points $x$ which work, so the part after the such that is vacuously true.

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    $\begingroup$ Often continuity is stated using $|x-c|\lt\delta$ instead of $0 < |x-c| < \delta$, which makes no logical difference, but in this case avoids appeal to vacuous truth. $\endgroup$ – Jonas Meyer Dec 27 '10 at 6:14
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    $\begingroup$ You are of course correct. It is interesting, and the OP should convince himself the two are equivalent because we require the limit to equal the value of the function. $\endgroup$ – TNi Dec 27 '10 at 6:24

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