Let $\tau_K$ be the $K$-topology on reals and $\tau$ the Euclidean topology on reals. Let $f\colon (\mathbb{R}, \tau_K) \to (\mathbb{R}, \tau)$ be a continous function. Is it the case that $f$ remains continuous when we interpret it as a function $f\colon (\mathbb{R}, \tau) \to (\mathbb{R}, \tau)$?
Topology $\tau_K$ consist of sets $U\setminus A$ where $U$ is open in Euclidean topology and $A \subset \{\,1/n : n \in \mathbb{Z}_+ \,\} =: K$. Hence $\tau \subsetneq \tau_K$. Check out Wikipedia for some interesting properties of $K$-topology.
I think the answer is yes. That is $C(\mathbb{R}, \tau_K) = C(\mathbb{R}, \tau)$. Below is the sketch of the proof. I will be glad for feedback on its correctness, alternative proof (mine seems like an overkill), or disproof/counterexample.
Context: $C(ℝ,τ_K) = C(ℝ)$ implies the weak topology on $ℝ$ generated by $C(ℝ,τ_K)$ is just the Euclidean topology.
Proof sketch. Assume for the sake of a contradiciton that $f\colon (\mathbb{R}, \tau_K) \to (\mathbb{R}, \tau)$ is continous, but $\hat f \colon (\mathbb{R}, \tau) \to (\mathbb{R}, \tau)$, which sends $x$ to $f(x)$, is not contionus.
Both $\tau_K$ and $\tau$ are separable, hence sequential spaces.
Therefore the continuity of $f$ is equivalent to saying that for any $\tau_K$-convergence sequence $x_n \to x$ we have $f(x_n) \to f(x)$.
And $\hat f$ is not continous iff there exist $\tau$-convergent sequence $x_n \to x$ for which $\hat f(x_n) \not \to \hat f(x)$.
Hence by assumption there is a sequence $(x_n)$ which is $\tau$-convergent to some $x$ (not $\tau_K$-convergent) and $\hat f(x_n) \not \to f(x)$.
But only $\tau$-convergent sequences which are not $\tau_K$-convergent are the ones $\tau$-converging to $0$ with infinitely many, decreasing terms of the form $1/m$. Let $(y_n)$ be such subsequence of $(x_n)$.
Take net $(z_\alpha)_{\alpha \in A}$ where $z_\alpha = \alpha \in A := (0,1) – K$ and $\preccurlyeq$ is the ordering of $A$ which is just reversed $\leq$. That is $a \preccurlyeq b \iff b \leq a$.
Net $(z_\alpha)$ conveges to $0$ in both topologies. Hence $f(x_\alpha) \to 0$ and $\hat f(\alpha) \to 0$.
Take $\epsilon> 0$ so small that infinitely many $\hat f(y_n)$ are in $U := (-\infty, -2\epsilon) \cup (2 \epsilon, +\infty)$. Such $\epsilon$ exists because $\hat f(y_n) \not \to \hat f(0)$. Let $B = B(\epsilon, f(0))$. On the other hand, because $f(x_\alpha) \to f(0)$, there is $\beta \in A$ such that $\beta \preccurlyeq \alpha$ implies $f(x_\alpha) \in B$. Hence for any $x \in (0, \beta) – K$ we have $f(x) \in B$.
The set
$$
V := f^{-1}(U) \cap f^{-1}(B) \cap (0, \beta)
$$
is $\tau_K$-open, becasue $f$ is $\tau_K$-contionus and $(0,\beta)$ is open. Note $(0,\beta) -K \subset f^{-1}(B)$. Set $V$ is not empty because infinitely many $y_n$'s are in $f^{-1}(U) \cap (0,\beta)$ and none of $y_n$ is in $(0, \beta)-K$, because $y_n \in K$.
But $V$ is a subset of $K$, which is not $\tau_K$ open — a contradiction.
Best Answer
Yes, $f$ remains continuous as a map $(\mathbb R,\tau)\to(\mathbb R,\tau)$. Here is an alternative approach, that avoids nets, sequences, and convergence.
It suffices to show that if $f(x)\in (a,b)$, then $f^{-1}((a,b))$ is a $\tau$-neighborhood of $x$. To see this, observe that there are $a',b'$ with $f(x)\in (a',b')\subset [a',b'] \subset (a,b)$. Then by $(\tau_K,\tau)$-continuity, we have
$$f^{-1}([a',b'])\supseteq f^{-1}((a',b'))\supseteq U\backslash K$$ for some $\tau$-open $U\ni x$. But $U\backslash K$ is dense in $U$ in the $\tau_K$ topology (since $K$ has no interior). Then, again by $(\tau_K,\tau)$-continuity, we have that $f^{-1}([a',b'])$ is closed, hence $$f^{-1}((a,b))\supseteq f^{-1}([a',b'])\supseteq \overline{ U\backslash K}\supset U\ni x.$$
Remark
The only two facts needed for this proof were that $K$ has no $\tau_K$-interior (which immediately follows from the fact that $K$ has no $\tau$-interior), and that $\mathbb R$ is regular, ie, given any point $x\in\mathbb R$ and any open set $U\ni x$ there is an open set $V$ with $$x\in V\subseteq \overline{V}\subseteq U.$$
As a result, the preceding proof generalizes if the $K$-topology is defined on any space $X$, with $K\subset X$ having empty interior, and $f\colon X\to Y$ is a map into a regular space $Y$.