Where $\mathbb{Z}$ is the set of integers and $\mathbb{R}$ the set of real numbers.

In a question in a problem sheet, it said this statement was correct, however I do not understand how.

You clearly cannot even begin to draw this function without a lot of gaps.

I suppose when the $\lim_{x\to Z_1} f(x) = f(Z_1)$. So is that the reason why the function is continuous?

Edit: This question came up in a first year university analysis module so I'm not too sure what topology means. Also, I use the standard epsilon, delta definition of continuity.

  • $\begingroup$ What is the function in question? $\endgroup$ – carmichael561 Jun 12 '16 at 20:17
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    $\begingroup$ It depends on the topology on $\mathbb{Z}$ and $\mathbb{R}$ $\endgroup$ – M10687 Jun 12 '16 at 20:17
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    $\begingroup$ How did you define "continuous"? $\endgroup$ – Milo Brandt Jun 12 '16 at 20:19
  • $\begingroup$ Continuity is, most generally, a topological notion. It's definition in terms of the topologies on the domain and codomain of $f$ is that the inverse image of any open set must be an open set. That is: $f^{-1}(V)=U$ where $V$ open in the codomain implies $U$ open in the domain. $\endgroup$ – Justin Benfield Jun 12 '16 at 20:23
  • $\begingroup$ Are you sure you can't draw it without a lot of "gaps"? $\endgroup$ – Neal Jun 12 '16 at 21:06

I'd assume you let $\mathbb Z$ have the discrete topology (where all one-point sets are open), and $\mathbb R$ have the usual topology.

Then a function $f: \mathbb Z \to \mathbb R$ is continous if the inverse image of an open set in $\mathbb R$ is open in $\mathbb Z$.

But every set is open in the discrete topology, because $U = \cup_{x \in U} \{ x \}$. Hence every function is continous.


Let me say like this, it $f:\Bbb Z \to \Bbb R$ is continuous since it cannot be noncontinuous.

Let me remind you, a function, $f$, is noncontinuous if there is an $x_0$ in its domain in which $f$ is noncontinuous in $x_0$.

By definition${}^\dagger$:

  • if a function $f$ is noncontinuous in $x_0$ then (there exists a small distance $\epsilon$, where for all neighborhoods of $x_0$ there are points, $p$, in which $|f(y)-f(x_0)|>\epsilon$).

But this couldn't happen since there are neighborhoods in which no or just a single point, $x_0$, exist. so $|f(p)-f(x_0)|=0.$

This denies the consequent of the of the previous proposition, so by modus tollens, $f$ is continuous.

${}^\dagger$ Maybe one may have not seen a definition for a function to be noncontinuous in a point, so he may define it as a negation to the definition of being continuous in a point. Since the definition of being continuous in a point is in the form of $\forall \epsilon \exists \delta P$, the definition of being noncontinuous in a point should be of the form of $\exists \epsilon \forall \delta \neg P$, with same $\epsilon,\delta,P$ as the previous definition.


This very much depends on what topologies you have assigned to $\mathbb{Z}$ and $\mathbb{R}$.

Let $D \subseteq \mathbb{R}$. The usual definition of $f:D \to \mathbb{R}$ being continuous at some $x_0 \in D$ (this assumes the usual topology) is that for each $\epsilon>0$ there is some $\delta>0$ such that for any $x\in D$ with $|x-x_0|<\delta$ we have $|f(x)-f(x_0)|<\epsilon$.

Take any function $f:\mathbb{Z}\to\mathbb{R}$. Pick some $x_0 \in \mathbb{Z}$ and let $\epsilon>0$. Then we need to come up with a $\delta$ to make the rest of the statement of the definition true. An easy choice is $\delta=1/2$.

Let $x \in \mathbb{Z}$ and suppose that $|x-x_0|<\delta=1/2$. Well, then since the only integer within 1/2 distance of $x_0$ is itself, we must have $x=x_0$. Thus $f(x)=f(x_0)$ and $|f(x)-f(x_0)|=0$ which is certainly less than $\epsilon$.

This shows that $f$ is continuous at $x_0$. Since $x_0$ was arbitrary, $f$ is continuous everywhere on its domain.

From the topological viewpoint, the subspace topology that $\mathbb{Z}$ inherits from $\mathbb{R}$ is the discrete topology. This means that every subset of $\mathbb{Z}$ is open (in $\mathbb{Z}$). The topological definition of continuity requires that the inverse image of an open set is an open set. So since all subsets of the domain are open, all inverse images of all subsets of the codomain (open or otherwise) are open. Thus every function is continuous.

Of course there is nothing special about $\mathbb{Z}$ and $\mathbb{R}$ here. If $X$ and $Y$ are topological spaces where $X$ has the discrete topology (i.e. all subsets are open), then all functions $f:X\to Y$ are continuous.


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