Study continuity of $f$ from $\mathbb{R}\times \mathbb{Z} \to \mathbb{R}$. 
Let $X=\mathbb{R}\times \mathbb{Z}$ and $\tau$ the product topology $\tau_u \times \tau_{CF}$ where $\tau_u$ is the usual topology on $\mathbb{R}$ and $\tau_{CF}$ the cofinite topology on $\mathbb{Z}$. Study the contiuity of $f : (X, \tau)\to (\mathbb{R},\tau_u)$ defined as
  $$(t,k)\mapsto t-k$$

What I tried was fixing the point $(t,k)=(0,0)$ and taking a general neighbourhood of $f(t,k)=0$ like $(-\varepsilon, \varepsilon)$ and a neighbourhood of $(0,0)$ like the product of $(-\delta, \delta)$ and $\mathbb{Z} \setminus \{1\}$ where $\varepsilon <\delta$. Then, take $r$ such that $\varepsilon < \delta -r$. Then
$$f(\delta -r, 0)=\delta -r > \varepsilon$$
so
$$f((-\delta, \delta) \times (\mathbb{Z}\setminus \{1\} )) \not \subset (-\varepsilon, \varepsilon)$$
and then $f$ is not continous.
But I'm not sure at all if this is right, so if you could help me with that I'd be grateful. Thanks in advance!
 A: You have the right general idea, but you’ve not executed it quite right. In order to show that $f$ is not continuous at $\langle 0,0\rangle$, it’s enough to find one open nbhd $U$ of $0$ such that $f^{-1}[U]$ is not open; let’s try $U=\left(-\frac12,\frac12\right)$. Clearly
$$\begin{align*}
f^{-1}[U]&=\left\{\langle t,k\rangle\in X:-\frac12<t-k<\frac12\right\}\\
&=\left\{\langle t,k\rangle\in X:k-\frac12<t<k+\frac12\right\}\;.
\end{align*}\tag{1}$$
For what values of $k$ is it true that $\langle 0,k\rangle\in f^{-1}[U]$? From $(1)$ we see that $\langle 0,k\rangle\in f^{-1}[U]$ if and only if
$$k-\frac12<0<k+\frac12\;,$$
and clearly this is true only for $k=0$. On the other hand, every open nbhd of $\langle 0,0\rangle$ in $X$ contains a basic open nbhd of the form $(-\epsilon,\epsilon)\times C$ for some $\epsilon>0$ and cofinite $C\subseteq\Bbb Z$ and therefore contains a set of the form $\{0\}\times C$ for some cofinite $C\subseteq\Bbb Z$. Thus, on the one hand we have
$$\{k\in\Bbb Z:\langle 0,k\rangle\in f^{-1}[U]\}=\{0\}\;,$$
and on the other hand we know that if $V$ is any open nbhd of $\langle 0,0\rangle$ in $X$, the set
$$\{k\in\Bbb Z:\langle 0,k\rangle\in V\}$$
is a cofinite subset of $\Bbb Z$ and hence in particular is infinite. It follows that $f^{-1}[U]$ does not contain any open nbhd of $\langle 0,0\rangle$ and hence that $f$ is not continuous at $\langle 0,0\rangle$.
It’s actually not hard to adapt this argument to show that $f$ is not continuous at any point of $X$.
A: Your idea is good, but you shouldn't consider a particular neighborhood of $(0,0)$.

Fix a point $(t_0,k_0)\in X$. A basic neighborhood of this point has the form $U=(t_0-\delta,t_0+\delta)\times C$, where $C$ is cofinite and $k_0\in C$, in particular both upper and lower unbounded.
Then
$$
f(U)=\bigcup_{k\in C}(t_0-k-\delta,t_0-k+\delta)
$$
which is unbounded, so it can't be contained in $(t_0-k_0-\varepsilon,t_0-k_0+\varepsilon)$, for any $\varepsilon>0$.
