In the following section, we introduce a new type of sequence spaces whose F̂ transform is I-convergent 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:
$$\begin{aligned} B_{x}(r,t) (\hat{F}) =&\{y=(y_{k})\in l_{\infty }:\\ &{} \bigl\{ k\in \mathbb{N}: \mu \bigl(\hat{F_{k}}(x)-\hat{F_{k}}(y),t \bigr)> 1-r\mbox{ and }\nu \bigl(\hat{F_{k}}(x)- \hat{F_{k}}(y),t \bigr)< r\bigr\} . \end{aligned}$$
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 non-empty. 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 right-hand 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:
$$\begin{aligned} B_{x}(r,t) (\hat{F}) =& \{y=(y_{k})\in l_{\infty }:\\ &{}\bigl\{ k\in \mathbb{N}: \mu \bigl(\hat{F_{k}}(x)- \hat{F_{k}}(y),t\bigr)> 1-r\mbox{ and }\nu \bigl(\hat{F_{k}}(x)- \hat{F_{k}}(y),t\bigr)< r\bigr\} . \end{aligned}$$
Let \(z=(z_{k})\in B_{x}(r,t)(\hat{F})\) so that \(\mu (\hat{F_{k}}(x)- \hat{F_{k}}(z),t)> 1-r\) and \(\nu (\hat{F_{k}}(x)-\hat{F_{k}}(z),t)< r\). Then there exists \(t_{0}\in (0,t)\) with \(\mu (\hat{F_{k}}(x)- \hat{F_{k}}(z),t_{0})>1-r \) 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 \(p_{0} > 1-r\), there exists \(s\in (0,1)\) such that \(p_{0}> 1-s> 1-r\). Using Remark 2.1(i), given \(p_{0}> 1-s\), we can find \(p_{1}, p_{2}\in (0,1)\) with \(p_{0}\ast p_{1} > 1-s\) and \((1-p_{0}) \diamond (1-p_{2})< s\). Put \(p_{3} = \max (p_{1}, p_{2})\). We will prove \(B_{z}(1- p_{3}, t-t_{0})(\hat{F})\subset B_{x}(r,t)(\hat{F})\). Let \(w=(w_{k})\in B_{z}(1-p_{3}, t - t_{0})(\hat{F})\). Hence
$$\begin{aligned}& \mu \bigl(\hat{F_{k}}(x)-\hat{F_{k}}(w),t\bigr)\geq \mu \bigl(\hat{F_{k}}(x)- \hat{F_{k}}(z),t_{0} \bigr)\ast \mu \bigl(\hat{F_{k}}(z)-\hat{F_{k}}(w),t-t_{0} \bigr) \\& \quad > (p_{0}\ast p_{3})\geq (p_{0}\ast p_{1}) > 1- s > 1 - r, \end{aligned}$$
and
$$\begin{aligned}& \nu \bigl(\hat{F_{k}}(x)-\hat{F_{k}}(w),t\bigr)\leq \nu \bigl(\hat{F_{k}}(x)- \hat{F_{k}}(z),t_{0} \bigr)\diamond \nu \bigl(\hat{F_{k}}(z)-\hat{F_{k}}(w),t-t _{0}\bigr) \\& \quad < (1-p_{0}))\diamond (1 - p_{3})\leq (1 - p_{0})\diamond (1 - p_{2}) < r. \end{aligned}$$
Hence \(w\in B_{x}(r,t)(\hat{F})\) and therefore \(B_{z}(1-p_{3},t- t _{0})(\hat{F})\subset B_{x}(r,t)(\hat{F})\). □
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 \(\mu (\hat{F_{k}}(x)-\xi ,t) > 1-r\) 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 \(x_{k}\rightarrow \xi \). □