In the following section, we introduce a new type of sequence spaces whose F̂ transform is Iconvergent with respect to the intuitionistic norm \((\mu ,\nu )\). Further we prove certain properties of these spaces such as hausdorfness, first countability. Throughout this paper, I is an admissible ideal. We define
$$\begin{aligned}& S_{0(\mu ,\nu )}^{I}(\hat{F})= \bigl\{ x=(x_{k})\in l_{\infty }:\bigl\{ k\in \mathbb{N}:\mu \bigl(\hat{F_{k}}(x),t \bigr)\leq 1\epsilon \mbox{ or }\nu \bigl( \hat{F_{k}}(x),t\bigr)\geq \epsilon \bigr\} \in I\bigr\} , \\& S_{(\mu ,\nu )}^{I}(\hat{F})=\bigl\{ x=(x_{k})\in l_{\infty }:\bigl\{ k\in \mathbb{N}:\mu \bigl(\hat{F_{k}}(x)l,t \bigr)\leq 1\epsilon \mbox{ or }\nu \bigl( \hat{F_{k}}(x)l,t\bigr) \geq \epsilon \bigr\} \in I\bigr\} . \end{aligned}$$
We introduce an open ball with centre x and radius r with respect to t as follows:
$$\begin{aligned} B_{x}(r,t) (\hat{F}) =&\{y=(y_{k})\in l_{\infty }:\\ &{} \bigl\{ k\in \mathbb{N}: \mu \bigl(\hat{F_{k}}(x)\hat{F_{k}}(y),t \bigr)> 1r\mbox{ and }\nu \bigl(\hat{F_{k}}(x) \hat{F_{k}}(y),t \bigr)< r\bigr\} . \end{aligned}$$
Remark 2.1

(i)
For \(p_{1} ,p_{2}\in (0,1)\) such that \(p_{1} > p_{2}\), there exist \(p_{3} , p_{4}\in (0,1)\) with \(p_{1}* p_{3} \geq p_{2}\) and \(p_{1}\geq p_{4} \diamond p_{2}\).

(ii)
For \(p_{5}\in (0,1)\), there exist \(p_{6} ,p_{7}\in (0,1)\) such that \(p_{6}\ast p_{6} \geq p_{5}\) and \(p_{7}\diamond p_{7} \leq p_{5}\).
Theorem 2.1
The spaces
\(S_{0(\mu ,\nu )}^{I}(\hat{F})\)
and
\(S_{(\mu , \nu )}^{I}(\hat{F})\)
are vector spaces over
\(\mathbb{R}\).
Proof
Let us show the result for \(S_{(\mu ,\nu )}^{I}(\hat{F})\) and the proof for another space will follow on the similar lines. Let \(x=(x_{k})\) and \(y=(y_{k})\in S_{(\mu ,\nu )}^{I}(\hat{F})\). Then by definition there exist \(\xi _{1}\) and \(\xi _{2}\), and for every ϵ, \(t > 0\), we have
$$\begin{aligned}& A=\biggl\{ k\in \mathbb{N}:\mu \biggl(\hat{F_{k}}(x)\xi _{1}, \frac{t}{2 \vert \alpha \vert }\biggr) \leq 1\epsilon \mbox{ or }\nu \biggl(\hat{F_{k}}(x)\xi _{1},\frac{t}{2 \vert \alpha \vert } \biggr)\geq \epsilon \biggr\} \in , \\& B=\biggl\{ k\in \mathbb{N}:\mu \biggl(\hat{F_{k}}(y)\xi _{2},\frac{t}{2 \vert \beta \vert }\biggr) \leq 1\epsilon \mbox{ or }\nu \biggl(\hat{F_{k}}(y)\xi _{2},\frac{t}{2 \vert \beta \vert } \biggr) \geq \epsilon \biggr\} \in I, \end{aligned}$$
where α and β are scalars.
$$\begin{aligned}& A^{c}=\biggl\{ k\in \mathbb{N}: \mu \biggl(\hat{F_{k}}(x) \xi _{1},\frac{t}{2 \vert \alpha \vert }\biggr) > 1\epsilon \mbox{ or }\nu \biggl(\hat{F_{k}}(x)\xi _{1}, \frac{t}{2 \vert \alpha \vert } \biggr) < \epsilon \biggr\} \in \mathcal{F}(I), \\& B^{c}=\biggl\{ k\in \mathbb{N}: \mu \biggl(\hat{F_{k}}(y) \xi _{2}, \frac{t}{2 \vert \beta \vert }\biggr) > 1\epsilon \mbox{ or }\nu \biggl(\hat{F_{k}}(y)\xi _{2},\frac{t}{2 \vert \beta \vert } \biggr) < \epsilon \biggr\} \in \mathcal{F}(I). \end{aligned}$$
Define \(E=A\cup B\) so that \(E\in I\). Thus \(E^{c}\in \mathcal{F}(I)\) and therefore is nonempty. We will show
$$\begin{aligned} E^{c} \subset& \bigl\{ k\in \mathbb{N}: \mu \bigl(\alpha \hat{F_{k}}(x)+ \beta \hat{F_{k}}(y)(\alpha \xi _{1}+\beta \xi _{2}),t\bigr)> 1\epsilon \mbox{ or}\\ &{}\nu \bigl(\alpha \hat{F_{k}}(x)+\beta \hat{F_{k}}(y)( \alpha \xi _{1}+ \beta \xi _{2}), t\bigr) < \epsilon \bigr\} . \end{aligned}$$
Let \(n\in E^{c}\). Then
$$\begin{aligned}& \mu \biggl(\hat{F_{n}}(x)\xi _{1}, \frac{t}{2 \vert \alpha \vert }\biggr) > 1\epsilon \mbox{ or } \nu \biggl( \hat{F_{n}}(x) \xi _{1}, \frac{t}{2 \vert \alpha \vert }\biggr)< \epsilon , \\& \mu \biggl(\hat{F_{n}}(y)\xi _{2}, \frac{t}{2 \vert \beta \vert }\biggr)> 1\epsilon \mbox{ or } \nu \biggl( \hat{F_{n}}(y)\xi _{2},\frac{t}{2 \vert \beta \vert }\biggr)< \epsilon . \end{aligned}$$
Consider
$$\begin{aligned}& \mu \bigl(\alpha \hat{F_{n}}(x)+ \beta \hat{F_{n}}(x)( \alpha \xi _{1}+ \beta \xi _{2}), t\bigr) \\& \quad \geq \mu \biggl(\alpha \hat{F_{n}}(x)\alpha \xi _{1},\frac{t}{2}\biggr) \ast \mu \biggl(\beta \hat{F_{n}}(y)\beta \xi _{2},\frac{t}{2} \biggr) \\& \quad = \mu \biggl(\hat{F_{n}}(x)\xi _{1}, \frac{t}{2 \vert \alpha \vert }\biggr) \ast \mu \biggl( \hat{F_{n}}(y)\xi _{2},\frac{t}{ \vert \beta \vert }\biggr) \\& \quad > (1\epsilon ) \ast (1\epsilon )=1\epsilon \end{aligned}$$
and
$$\begin{aligned}& \nu \bigl(\alpha \hat{F_{n}}(y)+ \beta \hat{F_{n}}(y)( \alpha \xi _{1}+ \beta \xi _{2})\bigr) \\& \quad \leq \nu \biggl(\alpha \hat{F_{n}}(x)\alpha \xi _{1},\frac{t}{2}\biggr) \diamond \nu \biggl(\beta \hat{F_{n}}(y)\beta \xi _{2},\frac{t}{2} \biggr) \\& \quad = \nu \biggl(\hat{F_{n}}(x)\xi _{1}, \frac{t}{2 \vert \alpha \vert }\biggr) \diamond \nu \biggl( \hat{F_{n}}(y)\xi _{2},\frac{t}{2 \vert \beta \vert }\biggr) \\& \quad < \epsilon \diamond \epsilon =\epsilon . \end{aligned}$$
Thus \(E^{c}\subset \{k\in \mathbb{N}: \mu (\alpha \hat{F_{k}}(x)+ \beta \hat{F_{k}}(y)(\alpha \xi _{1}+ \beta \xi _{2}),t)>1\epsilon \mbox{ or }\nu (\alpha \hat{F_{k}}(x)+\beta \hat{F_{k}}(y)(\alpha \xi _{1} + \beta \xi _{2}),t)< \epsilon \}\). \(E^{c}\in \mathcal{F}(I)\), therefore by definition of filter, the set on the righthand side of the above equation belongs to \(\mathcal{F}(I)\) so that its complement belongs to I. This implies \((\alpha x+ \beta y)\in S_{(\mu , \nu )}^{I}(\hat{F})\). Hence \(S_{(\mu ,\nu )}^{I}(\hat{F})\) is a vector space over \(\mathbb{R}\). □
Theorem 2.2
Every open ball
\(B_{x}(r,t)(\hat{F})\)
is an open set in
\(S_{(\mu ,\nu )}^{I}(\hat{F})\).
Proof
We have defined open ball as follows:
$$\begin{aligned} B_{x}(r,t) (\hat{F}) =& \{y=(y_{k})\in l_{\infty }:\\ &{}\bigl\{ k\in \mathbb{N}: \mu \bigl(\hat{F_{k}}(x) \hat{F_{k}}(y),t\bigr)> 1r\mbox{ and }\nu \bigl(\hat{F_{k}}(x) \hat{F_{k}}(y),t\bigr)< r\bigr\} . \end{aligned}$$
Let \(z=(z_{k})\in B_{x}(r,t)(\hat{F})\) so that \(\mu (\hat{F_{k}}(x) \hat{F_{k}}(z),t)> 1r\) and \(\nu (\hat{F_{k}}(x)\hat{F_{k}}(z),t)< r\). Then there exists \(t_{0}\in (0,t)\) with \(\mu (\hat{F_{k}}(x) \hat{F_{k}}(z),t_{0})>1r \) and \(\nu (\hat{F_{k}}(x)\hat{F_{k}}(z),t _{0})< r\). Put \(p_{0}= \mu (\hat{F_{k}}(x)\hat{F_{k}}(z),t_{0})\), so we have \(p_{0} > 1r\), there exists \(s\in (0,1)\) such that \(p_{0}> 1s> 1r\). Using Remark 2.1(i), given \(p_{0}> 1s\), we can find \(p_{1}, p_{2}\in (0,1)\) with \(p_{0}\ast p_{1} > 1s\) and \((1p_{0}) \diamond (1p_{2})< s\). Put \(p_{3} = \max (p_{1}, p_{2})\). We will prove \(B_{z}(1 p_{3}, tt_{0})(\hat{F})\subset B_{x}(r,t)(\hat{F})\). Let \(w=(w_{k})\in B_{z}(1p_{3}, t  t_{0})(\hat{F})\). Hence
$$\begin{aligned}& \mu \bigl(\hat{F_{k}}(x)\hat{F_{k}}(w),t\bigr)\geq \mu \bigl(\hat{F_{k}}(x) \hat{F_{k}}(z),t_{0} \bigr)\ast \mu \bigl(\hat{F_{k}}(z)\hat{F_{k}}(w),tt_{0} \bigr) \\& \quad > (p_{0}\ast p_{3})\geq (p_{0}\ast p_{1}) > 1 s > 1  r, \end{aligned}$$
and
$$\begin{aligned}& \nu \bigl(\hat{F_{k}}(x)\hat{F_{k}}(w),t\bigr)\leq \nu \bigl(\hat{F_{k}}(x) \hat{F_{k}}(z),t_{0} \bigr)\diamond \nu \bigl(\hat{F_{k}}(z)\hat{F_{k}}(w),tt _{0}\bigr) \\& \quad < (1p_{0}))\diamond (1  p_{3})\leq (1  p_{0})\diamond (1  p_{2}) < r. \end{aligned}$$
Hence \(w\in B_{x}(r,t)(\hat{F})\) and therefore \(B_{z}(1p_{3},t t _{0})(\hat{F})\subset B_{x}(r,t)(\hat{F})\). □
Remark 2.2
Let \(S_{(\mu ,\nu )}^{I}(\hat{F}\) be IFNS. Define \(\tau _{(\mu ,\nu )} ^{I}(\hat{F})=\{A\subset S_{(\mu ,\nu )}^{I}(\hat{F})\mbox{: for given } x\in A\), we can find \(t>0\) and \(0 < r <1\) such that \(B_{x}(r,t)( \hat{F})\subset A\}\). Then \(\tau _{(\mu ,\nu )}^{I}(\hat{F})\) is a topology on \(S_{(\mu ,\nu )}^{I}(\hat{F})\).
Remark 2.3
Since \(\{B_{x}(\frac{1}{n},\frac{1}{n})(\hat{F}): n\in \mathbb{N}\}\) is a local base at x, the topology \(\tau _{(\mu ,\nu )}^{I}(\hat{F})\) is first countable.
Theorem 2.3
The spaces
\(S_{(\mu ,\nu )}^{I}(\hat{F})\)
and
\(S_{0(\mu , \nu )}^{I}(\hat{F})\)
are Hausdorff.
Proof
Let \(x,y\in S_{(\mu ,\nu )}^{I}(\hat{F})\) with x and y to be different. Then \(0< \mu (\hat{F_{k}}(x)\hat{F}(y),t) <1\) and \(0<\nu (\hat{F}(x)\hat{F}_{k}(y),t)< 1\). Put \(\mu (\hat{F_{k}}(x) \hat{F}_{k}(y),t) = p_{1}\) and \(\nu (\hat{F}_{k}(x)\hat{F}{k}(y),t) = p_{2}\) and \(r = \max (p_{1}, 1 p_{2})\). Using Remark (2.1(ii)) for \(p_{0}\in (r, 1)\), we can find \(p_{3}, p_{4}\in (0,1)\) such that \(p_{3} \ast p_{3} \geq p_{0}\) and \((1 p_{4})\diamond (1 p_{4}) \leq 1 p_{0}\). Put \(p_{5} = \max (p_{3}, p_{4})\). Clearly \(B_{x}(1 p _{5},\frac{t}{2})(\hat{F})\cap B_{y}(1 p_{5},\frac{t}{2})(\hat{F})= \phi \). Let on the contrary \(z\in B_{x}(1  p_{5},\frac{t}{2})( \hat{F})\cap B_{y}(1  p_{5},\frac{t}{2})(\hat{F})\). Then we have
$$\begin{aligned} p_{1} =& \mu \bigl(\hat{F_{k}}(x)\hat{F_{k}}(y),t \bigr) \geq \mu \biggl(\hat{F_{k}}(x) \hat{F_{k}}(z), \frac{t}{2}\biggr) \ast \mu \biggl(\hat{F_{k}}(z) \hat{F_{k}}(y), \frac{t}{2}\biggr) \\ \geq& p_{5} \ast p_{5} \geq p_{3} \ast p_{3}> p_{0} > p_{1} \end{aligned}$$
and
$$\begin{aligned} p_{2} =& \nu \bigl(\hat{F_{k}}(x) \hat{F_{y}},t\bigr) \leq \nu \biggl(\hat{F_{k}}(x) \hat{F_{k}}(z),\frac{t}{2}\biggr) \diamond \nu \biggl( \hat{F_{k}}(z)\hat{F_{k}}(y), \frac{t}{2} \biggr) \\ \leq& (1 p_{5})\diamond (1 p_{5}) \leq (1 p_{4})\diamond (1 p_{4}) \leq 1 p_{0} < p_{2}, \end{aligned}$$
which is a contradiction. Therefore \(S_{(\mu ,\nu )}^{I}(\hat{F})\) is a Hausdorff space. The proof for \(S_{0(\mu ,\nu )}^{I}(\hat{F})\) follows similarly. □
Theorem 2.4
Let
\(S_{(\mu ,\nu )}^{I}(\hat{F})\)
be IFNS and
\(\tau _{(\mu , \nu )}^{I}(\hat{F})\)
be a topology on
\(S_{\mu ,\nu )}^{I}(\hat{F})\). A sequence
\((x_{k})\in S_{(\mu ,\nu )}^{I}(\hat{F})\)
converges to
ξ
iff
\(\mu (\hat{F_{k}}(x)\xi ,t) \rightarrow 1\)
and
\(\nu ( \hat{F_{k}}(x)\xi ,t) \rightarrow 0\)
as
\(k\rightarrow \infty \).
Proof
Suppose \(x_{k}\rightarrow \xi \), then given \(0< r < 1\) there exists \(k_{0}\in \mathbb{N}\) such that \((x_{k})\in B_{x}(r,t)(\hat{F})\) for all \(k\geq k_{0}\) given \(t>0\). Hence, we have \(1\mu (\hat{F_{k}}(x) \xi ,t)< r\) and \(\nu (\hat{F_{k}}(x)\xi,t)< r\). Therefore \(\mu (\hat{F_{k}}(x)\xi,t)\rightarrow 1\) and \(\nu ( \hat{F_{k}}(x)\xi ,t)\rightarrow 0\) as \(k \rightarrow \infty \).
Conversely, if \(\mu (\hat{F_{k}}(x)\xi ,t)\rightarrow 1\) and \(\nu (\hat{F_{k}}(x)\xi ,t)\rightarrow 0\) as \(k\rightarrow \infty \) holds for each \(t>0\). For \(0< r< 1\), there exists \(k_{0}\in \mathbb{N}\) such that \(1\mu (\hat{F_{k}}(x)\xi ,t)< r\) and \(\nu (\hat{F_{k}}(x)\xi ,t)< r\) for all \(k\geq k_{0}\), which implies \(\mu (\hat{F_{k}}(x)\xi ,t) > 1r\) and \(\nu (\hat{F_{k}}(x)\xi ,t) < r\). Thus \(x_{k}\in B_{x}(r,t)(\hat{F})\) for all \(k\geq k_{0}\) and hence \(x_{k}\rightarrow \xi \). □