Definition 1 (see [17], [18].)Let W and D be two metric spaces, the set-valued mapping be expressed as ℵ : W⟶P₀(D), where P₀(D) is the nonempty set, ℵ is said to be

• (1)

  lower semicontinuous (l.s.c.) at w ∈ W if for any open set ℧ in D, ℵ(w)∩℧≠∅, there exists O(w) of w, where O(w) is an open neighbourhood such that ℵ(w^′)∩℧≠∅, for every w^′ ∈ O(w)

• (2)

  upper semi-continuous (u.s.c.) at w ∈ W if for any open set ℧ in D, ℵ(w) ⊂ ℧, there exists O(w) of w, where O(w) is an open neighbourhood such that ℵ(w^′) ⊂ ℧, for every w^′ ∈ O(w)

• (3)

  a usco mapping on W if for any w ∈ W, ℵ : W⟶P₀(D) is u.s.c and nonempty compact values

• (4)

  continuous at w ∈ W if ℵ is both u.s.c and l.s.c. at w

Lemma 2 (see [19].)Let {A_m} be a sequence of nonempty bounded subset of Rⁿ and A represents a nonempty bounded subset of Rⁿ, where for each m = 1, 2, 3, ⋯. If A_m⟶A and t_m ∈ A_m, then there exists a subsequence $$ of {t_m} such that $$.

Lemma 3 (see [19].)Let {A_m} denote a sequence of nonempty bounded subset of Rⁿ and A represents a nonempty bounded subset of Rⁿ, where for each m = 1, 2, 3, ⋯. Denote by G an open set of Rⁿ, where G ⋂ A ≠ ∅. If A_m⟶A, thus there exists a positive integer N such that for each m ≥ N, G ⋂ A_m ≠ ∅(m⟶∞).

Definition 4 (see [1].)Let ξ : Rⁿ⟶Rⁿ represent a mapping, then ξ is monotonic on Rⁿ if for any ς, σ ∈ Rⁿ, one has

Definition 5 (see [1].)Assume that two mappings are denoted as ξ, ϕ : Rⁿ⟶Rⁿ, respectively. If there exists a constant δ > 0 such that for every ς, σ ∈ Rⁿ, one has

Lemma 6 (see [17].)Let M and F be two topological spaces, where F be a compact space. If the set-valued mapping ℵ : M⟶P₀(F) is closed, then ℵ is a usco mapping on M.

Definition 7 (see [20].)Let ξ : E⟶R and E denotes a nonempty convex set of Rⁿ, then

• (1)

  ξ is quasiconvex on E, if for all e₁, e₂ ∈ E and ρ ∈ (0, 1), we have

• (2)

  ξ is quasiconcave on E, if for all e₁, e₂ ∈ E and ρ ∈ (0, 1), we have

Lemma 8 (see [21], Fort Lemma.)Let M be a Hausdorff topological space and W be a metric space. The set-valued mapping ℵ : M⟶P₀(W) is a usco mapping. Then, there exists a residual subset Q of M such that ℵ is l.s.c. on Q.

Definition 9. Let S be a nonempty compact subset of Rⁿ and denote by T(S) a set of all nonempty compact subsets in S. Let φ, ψ : S⟶S represent two continuous mappings. A set-valued mapping represents as H : S⟶T(S) such that for any s ∈ S, H(s) is a nonempty convex compact set. For a real number ε > 0, finding a vector s^∗ ∈ S such that ψ(s^∗) ∈ H(s^∗), we have

Then, s^∗ is said to be an ε-approximate solution of IQVI problems.

Theorem 10. Let S be a nonempty compact subset of Rⁿ and satisfy the following assumptions:

• (i)

  For every m = 1, 2, ⋯, the two function sequences φ_m, ψ_m : S⟶S and a set-valued mapping sequence H_m : S⟶T(S) are satisfied by

  $$ (7)

• (ii)

  For every m = 1, 2, ⋯, A_m is a nonempty subset sequence of S and

  $$ (8)

• (iii)

  For every m = 1, 2, ⋯, s_m ∈ S is satisfied with d(s_m, A_m)⟶0, ψ_m(s_m) ∈ H_m(s_m),we have

  $$ (9)

• (1)

  Then there exists a convergent subsequence $$ of {s_m} which converges to some s^∗ ∈ A(m⟶∞)

• (2)

  For all h ∈ H(s^∗), we have

  $$ (10)

• (3)

  If the solution of IQVI problems is a singleton set, there must be s_m⟶s^∗

Proof.

• (1)

  Since (s_m, A_m)⟶0(m⟶∞), we can see that there exists s_m′ ∈ A_m such that d(s_m, s_m′)⟶0. Because h(A_m, A)⟶0 and A is a compact set by Lemma 2, for any sequence {s_m′} in S, there must be a subsequence $$ such that $$. Therefore, there exists a subsequence $$ of the sequence {s_m} such that $$

• (2)

  According to conclusion (1), it may be assumed that s_m⟶s^∗ ∈ A. By contradiction, we assume that conclusion (2) does not hold. Thus, there exists h ∈ H(s^∗) such that

First, since φ, ψ : S⟶S are continuous at s and 〈φ(s), h − ψ(s)〉 is continuous at variables s and h, there exists a real number γ₀ > 0 and two open neighbourhoods V(s^∗) and V(h⁰) of s^∗ and h⁰ such that for any s^′ ∈ V(s^∗), h^′ ∈ V(h⁰), we have

According to the Cauchy-Swartz inequality, we can obtain

This means that

Again, because ε_m⟶0(m⟶∞), there exists a positive integer N₁ such that for every m ≥ N₁, we have

Finally, since s_m⟶s^∗ ∈ A, h(H_m, H)⟶0(m⟶∞), and h⁰ ∈ H(s), according to Lemma 3, we know that there exists a positive integer N₂. It may be assumed that N₂ ≥ N₁ such that for all m ≥ N₂, we have s_m ∈ V(s^∗) and V(h⁰) ⋂ H_m(s_m) ≠ ∅. Let h_m ∈ V(h⁰) ⋂ H_m(s_m), thus, we have

Then,

Then, this conflicts with condition (iii). Hence, for all h ∈ H(s^∗), we have 〈φ(s^∗), h − ψ(s^∗)〉 ≥ 0.

• (3)

  First, using the contradiction method, we assume that conclusion (3) does not hold. In other words, if the solution set of IQVI problems is a singleton set, thens_m↛s^∗. Hence, there exists γ > 0 and a subsequence $$ of {s_m} such that $$. Next, by conclusion (1), it can be seen that sequence $$ must have subsequences. Let $$, that is, $$, according to conclusion (2), we can obtain $$. Finally, because the solution of IQVI problems is singleton set, therefore $$, that is, $$, which conflicts with $$. Therefore, we can obtain s_m⟶s^∗. The proof is completed.

Remark 11. According to Theorem 10, although the objective function, feasible solution set are all approximated (that is, φ_m⟶φ, ψ_m⟶ψ, H_m⟶H, and A_m⟶A), we can obtain an approximation sequence {s_m}, which must have convergent subsequences $$, that is, $$ and s^∗ must be the solution of IQVI problems. If the ε_m-approximate solution s_m of IQVI problems is regarded as a “satisfactory solution” under bounded rationality, and the solution s^∗ of IQVI problems is regarded as an“exact solution” under full rationality. Theorem 10 implies the approximate of bounded rationality to full rationality, that is, full rationality can be approximated by a series of approximate solutions of bounded rationality, which verifies Simon’s bounded rationality theory from a certain perspective.

Remark 12. Theorem 10 shows that the limit points and convergent subsequences existing for sequence {s_m} are equivalent. It can be seen from the above proof that the sequence {s_m} must have limit points. Each limit point belongs to compact set A and is the solution of IQVI problems. If the solution of IQVI problems is a singleton set, there is a stronger convergence result as follows: s_m⟶s.

Corollary 13. Let S be a nonempty compact subset of Rⁿ and satisfy the following assumptions:

• (i)

  For every m = 1, 2, ⋯, the two function sequences φ_m, ψ_m : S⟶S and a set-valued mapping sequence H_m : S⟶T(S) are satisfied by

• (ii)

  For every m = 1, 2, ⋯, A_m is a nonempty subset of S and

  $$ (20)

• (iii)

  For every m = 1, 2, ⋯, s_m ∈ A_m is satisfied with ψ_m(s_m) ∈ H_m(s_m), we have

Corollary 14. Let S be a nonempty compact subset of Rⁿ and all the following assumptions be satisfied:

• (i)

  For every m = 1, 2, ⋯, the two function sequences φ_m, ψ_m : S⟶S and a set-valued mapping sequence H_m : S⟶T(S) are satisfied by

• (ii)

  A is a nonempty compact set of S:

• (iii)

  For every m = 1, 2, ⋯, s_m ∈ A_m is an ε_m-approximate solution of function sequences for the IQVI problem, which satisfies ψ_m(s_m) ∈ H_m(s_m), we have

  $$ (24)

Theorem 15. (U, ℓ) is a complete metric space.

Proof. Clearly, (U, ℓ) is a metric space. Therefore, we only need to prove that (U, ℓ) is complete.

First, we assume that any Cauchy sequence in U is $$. This means that for every θ > 0, there exists an integer N₁ such that any p, m ≥ N₁, we have

Then, there exist φ, ψ : S⟶S and H : S⟶T(S) such that $$, $$, we can obtain

Because s_m⟶s, φ_m⟶φ, and ψ_m⟶ψ, it is clear that φ and ψ are continuous at s and H(s) is a convex set. Since H_m(s) is the Cauchy sequence in T(S). According to a theorem in [20], (T(S), h) is the complete metric space. Therefore, there exists H : S⟶T(S) such that H_m(s)⟶H(s), for any s ∈ S, H(s) is a nonempty compact set, then H(s) is a nonempty convex compact set and H is continuous at s.

Next, we verify that φ (or ψ) is monotone quasiconcave (or monotone quasiconvex) and (φ, ψ) is a strongly monotone couple. Since φ_m, ψ_m are monotone, (φ_m, ψ_m) is a strongly monotone couple, and φ, ψ are continuous at s, that is, φ_m⟶φ, ψ_m⟶ψ, then φ, ψ are monotone and (φ, ψ) is a strongly monotone couple. Again, because φ_m (or ψ_m) is quasiconcave (or quasiconvex), for every s₁, s₂ ∈ S, ρ ∈ (0, 1), we have

Since φ, ψ are continuous at s, then there exists a real parameter θ > 0 such that

Therefore, we can obtain

By the arbitrariness of θ, then φ is quasiconcave and ψ is quasiconvex at s.

Again, according to u_m = (φ_m, ψ_m, H_m) ∈ U, there exists s_m ∈ S such that ψ_m(s_m) ∈ H_m(s_m), we have

Since S is a compact set, there exists s ∈ S such that s is the convergence point of {s_m}. It may be assumed that s_m⟶s and H_m(s)⟶H(s), H is continuous at s, then

Because ψ is continuous at s, ψ_m(s)⟶ψ(s), we define the distance on S is d, then

Since H(s) is a nonempty convex compact set, we can obtain the result that ψ(s) ∈ H(s).

Finally, by the contradiction method, we suppose that there exists h⁰ ∈ H(s) satisfying 〈φ(s), h⁰ − ψ(s)〉 < 0; thus, there exists a small enough β > 0 such that

Because h(H_m(s_m), H(s))⟶0 and h⁰ ∈ H(s), there exists h_m ∈ H_m(s_m) such that h_m⟶h⁰. Since φ(·), ψ(·) are continuous at s and φ_m⟶φ, ψ_m⟶ψ, s_m⟶s, h_m⟶h⁰, there exists m₁ ≥ 0 such that for any m ≥ m₁, we can obtain

Then, this conflicts with (33). So u = (φ, ψ, H) ∈ U. Therefore, the metric space (U, ℓ) is complete. The proof is completed.

Lemma 16. Ω is a usco mapping on U.

Proof. Since S be a nonempty compact subset, according to Lemma 6, we only need to prove that Ω is closed, which means that we have to prove that Graph(Ω) is closed, that is, for each u_m ∈ U, u_m⟶u and for each s_m ∈ Ω(u_m) and s_m⟶s, then s ∈ Ω(u).

For all m = 1, 2, ⋯, since s_m ∈ Ω(u_m), we have

Since φ(·), ψ(·) are continuous at s and φ_m⟶φ, ψ_m⟶ψ, s_m⟶s, therefore,

Then, we can obtain φ_m(s_m)⟶φ(s), ψ_m(s_m)⟶ψ(s). Since h − ψ_m(s_m)⟶h − ψ(s), we have

Thus, s ∈ Ω(u). The proof is completed.

Theorem 17. There exists a dense residual subset Q of U such that for all u = (φ, ψ, H) ∈ Q, Ω(u) is a singleton set.

Proof. First, because U is a complete metric space and Ω is a usco mapping on U, by the Fort lemma, there exists a dense residual subset Q of U such that for all u = (φ, ψ, H) ∈ Q, the set-valued mapping Ω is l.s.c. at u.

For all u = (φ, ψ, H) ∈ Q, we suppose that Ω(u) is not a singleton set; thus, there exist s₁, s₂ ∈ Ω(u), where s₁ ≠ s₂. Simultaneously, there exist two open neighbourhoods K of s₁ and L of s₂, respectively, such that K ⋂ L ≠ ∅ and K ⋂ Ω(u) ≠ ∅. Let s₁ ∈ K ⋂ Ω(u), then for all h ∈ H(s₁), we can obtain

For each s ∈ S, m = 1, 2, ⋯, we define

Therefore, it is easy to see that φ_m, ψ_m is continuous.

Next, because φ, ψ are monotone and for all s, l ∈ S, by Definition 4, we can obtain

Similarly,

Since (φ, ψ) is a strongly monotone couple on S, by Definition 5, there exists a constant δ > 0 and for each s, l ∈ S, we have

Then,

Therefore, (φ_m, ψ_m) is a strongly monotone couple on S.

Again, because φ is quasiconcave and ψ is quasiconvex, so for all ρ ∈ (0, 1), according to Definition 7 we can obtain

Similarly,

Then, φ_m is quasiconcave and ψ_m is quasiconvex.

Finally, for all h ∈ H_m(s₁), we have

Therefore, s₁ ∈ Ω(u_m), u_m ∈ U. Obviously, ℓ(u_m, u)⟶0(m⟶∞).

Note that s₂ ∈ L, then L ⋂ Ω(u) ≠ ∅. Since the set-valued mapping Ω is l.s.c. at u, there exists a sufficiently large integer m₀ such that $$. Take $$, we have

Let h = ψ(s₁), then

Because s₁ ∈ Ω(u), for every h ∈ H(s₁), we have

Let $$, then

This contradicts with the fact that (φ, ψ) is a strongly monotone couple on S. Therefore, for any u = (φ, ψ, H) ∈ Q, Ω(u) is a singleton set. This completes the proof.

Theorem 18. There exists a dense residual subset Q of U such that for all u = (φ, ψ, H) ∈ Q, there must be s_m⟶s, ψ(s) ∈ H(s), and 〈φ(s), h − ψ(s)〉 ≥ 0 for each h ∈ H(s).

Proof. According to Theorem 17, there exists a dense residual subset Q of U such that for any u = (φ, ψ, H) ∈ Q, the solution of the IQVI problem is a singleton set. By conclusion (3) of Theorem 10, there must be s_m⟶s. This completes the proof.

Remark 19. Theorem 18 shows that the solution set of the IQVI problem has generic convergence in the case of perturbation with the objective function on S.