Article TSP_1, MML version 4.99.1005
:: TSP_1:modenot 1 => TSP_1:mode 1
definition
let a1 be TopStruct;
redefine mode SubSpace of A1 -> TopStruct means
the carrier of it c= the carrier of a1 &
(for b1 being Element of bool the carrier of it holds
b1 is open(it)
iff
ex b2 being Element of bool the carrier of a1 st
b2 is open(a1) & b1 = b2 /\ the carrier of it);
end;
:: TSP_1:dfs 1
definiens
let a1, a2 be TopStruct;
To prove
a2 is SubSpace of a1
it is sufficient to prove
thus the carrier of a2 c= the carrier of a1 &
(for b1 being Element of bool the carrier of a2 holds
b1 is open(a2)
iff
ex b2 being Element of bool the carrier of a1 st
b2 is open(a1) & b1 = b2 /\ the carrier of a2);
:: TSP_1:def 1
theorem
for b1, b2 being TopStruct holds
b2 is SubSpace of b1
iff
the carrier of b2 c= the carrier of b1 &
(for b3 being Element of bool the carrier of b2 holds
b3 is open(b2)
iff
ex b4 being Element of bool the carrier of b1 st
b4 is open(b1) & b3 = b4 /\ the carrier of b2);
:: TSP_1:th 2
theorem
for b1 being TopStruct
for b2 being SubSpace of b1
for b3 being Element of bool the carrier of b1
st b3 is open(b1)
holds ex b4 being Element of bool the carrier of b2 st
b4 is open(b2) & b4 = b3 /\ the carrier of b2;
:: TSP_1:modenot 2 => TSP_1:mode 2
definition
let a1 be TopStruct;
redefine mode SubSpace of A1 -> TopStruct means
the carrier of it c= the carrier of a1 &
(for b1 being Element of bool the carrier of it holds
b1 is closed(it)
iff
ex b2 being Element of bool the carrier of a1 st
b2 is closed(a1) & b1 = b2 /\ the carrier of it);
end;
:: TSP_1:dfs 2
definiens
let a1, a2 be TopStruct;
To prove
a2 is SubSpace of a1
it is sufficient to prove
thus the carrier of a2 c= the carrier of a1 &
(for b1 being Element of bool the carrier of a2 holds
b1 is closed(a2)
iff
ex b2 being Element of bool the carrier of a1 st
b2 is closed(a1) & b1 = b2 /\ the carrier of a2);
:: TSP_1:def 2
theorem
for b1, b2 being TopStruct holds
b2 is SubSpace of b1
iff
the carrier of b2 c= the carrier of b1 &
(for b3 being Element of bool the carrier of b2 holds
b3 is closed(b2)
iff
ex b4 being Element of bool the carrier of b1 st
b4 is closed(b1) & b3 = b4 /\ the carrier of b2);
:: TSP_1:th 4
theorem
for b1 being TopStruct
for b2 being SubSpace of b1
for b3 being Element of bool the carrier of b1
st b3 is closed(b1)
holds ex b4 being Element of bool the carrier of b2 st
b4 is closed(b2) & b4 = b3 /\ the carrier of b2;
:: TSP_1:attrnot 1 => T_0TOPSP:attr 2
notation
let a1 be TopStruct;
synonym T_0 for discerning;
end;
:: TSP_1:attrnot 2 => T_0TOPSP:attr 2
definition
let a1 be TopStruct;
attr a1 is T_0 means
(a1 is not empty) implies for b1, b2 being Element of the carrier of a1
st b1 <> b2 &
(for b3 being Element of bool the carrier of a1
st b3 is open(a1) & b1 in b3
holds b2 in b3)
holds ex b3 being Element of bool the carrier of a1 st
b3 is open(a1) & not b1 in b3 & b2 in b3;
end;
:: TSP_1:dfs 3
definiens
let a1 be TopStruct;
To prove
a1 is discerning
it is sufficient to prove
thus (a1 is not empty) implies for b1, b2 being Element of the carrier of a1
st b1 <> b2 &
(for b3 being Element of bool the carrier of a1
st b3 is open(a1) & b1 in b3
holds b2 in b3)
holds ex b3 being Element of bool the carrier of a1 st
b3 is open(a1) & not b1 in b3 & b2 in b3;
:: TSP_1:def 3
theorem
for b1 being TopStruct holds
b1 is discerning
iff
(b1 is empty or for b2, b3 being Element of the carrier of b1
st b2 <> b3 &
(for b4 being Element of bool the carrier of b1
st b4 is open(b1) & b2 in b4
holds b3 in b4)
holds ex b4 being Element of bool the carrier of b1 st
b4 is open(b1) & not b2 in b4 & b3 in b4);
:: TSP_1:attrnot 3 => T_0TOPSP:attr 2
definition
let a1 be TopStruct;
attr a1 is T_0 means
(a1 is not empty) implies for b1, b2 being Element of the carrier of a1
st b1 <> b2 &
(for b3 being Element of bool the carrier of a1
st b3 is closed(a1) & b1 in b3
holds b2 in b3)
holds ex b3 being Element of bool the carrier of a1 st
b3 is closed(a1) & not b1 in b3 & b2 in b3;
end;
:: TSP_1:dfs 4
definiens
let a1 be TopStruct;
To prove
a1 is discerning
it is sufficient to prove
thus (a1 is not empty) implies for b1, b2 being Element of the carrier of a1
st b1 <> b2 &
(for b3 being Element of bool the carrier of a1
st b3 is closed(a1) & b1 in b3
holds b2 in b3)
holds ex b3 being Element of bool the carrier of a1 st
b3 is closed(a1) & not b1 in b3 & b2 in b3;
:: TSP_1:def 4
theorem
for b1 being TopStruct holds
b1 is discerning
iff
(b1 is empty or for b2, b3 being Element of the carrier of b1
st b2 <> b3 &
(for b4 being Element of bool the carrier of b1
st b4 is closed(b1) & b2 in b4
holds b3 in b4)
holds ex b4 being Element of bool the carrier of b1 st
b4 is closed(b1) & not b2 in b4 & b3 in b4);
:: TSP_1:condreg 1
registration
cluster non empty trivial -> discerning (TopStruct);
end;
:: TSP_1:condreg 2
registration
cluster non empty non discerning -> non trivial (TopStruct);
end;
:: TSP_1:exreg 1
registration
cluster non empty strict TopSpace-like discerning TopStruct;
end;
:: TSP_1:exreg 2
registration
cluster non empty strict TopSpace-like non discerning TopStruct;
end;
:: TSP_1:condreg 3
registration
cluster non empty TopSpace-like discrete -> discerning (TopStruct);
end;
:: TSP_1:condreg 4
registration
cluster non empty TopSpace-like non discerning -> non discrete (TopStruct);
end;
:: TSP_1:condreg 5
registration
cluster non empty non trivial TopSpace-like anti-discrete -> non discerning (TopStruct);
end;
:: TSP_1:condreg 6
registration
cluster non empty TopSpace-like anti-discrete discerning -> trivial (TopStruct);
end;
:: TSP_1:condreg 7
registration
cluster non empty non trivial TopSpace-like discerning -> non anti-discrete (TopStruct);
end;
:: TSP_1:attrnot 4 => T_0TOPSP:attr 2
definition
let a1 be TopStruct;
attr a1 is T_0 means
for b1, b2 being Element of the carrier of a1
st b1 <> b2
holds Cl {b1} <> Cl {b2};
end;
:: TSP_1:dfs 5
definiens
let a1 be non empty TopSpace-like TopStruct;
To prove
a1 is discerning
it is sufficient to prove
thus for b1, b2 being Element of the carrier of a1
st b1 <> b2
holds Cl {b1} <> Cl {b2};
:: TSP_1:def 5
theorem
for b1 being non empty TopSpace-like TopStruct holds
b1 is discerning
iff
for b2, b3 being Element of the carrier of b1
st b2 <> b3
holds Cl {b2} <> Cl {b3};
:: TSP_1:attrnot 5 => T_0TOPSP:attr 2
definition
let a1 be TopStruct;
attr a1 is T_0 means
for b1, b2 being Element of the carrier of a1
st b1 <> b2 & b1 in Cl {b2}
holds not b2 in Cl {b1};
end;
:: TSP_1:dfs 6
definiens
let a1 be non empty TopSpace-like TopStruct;
To prove
a1 is discerning
it is sufficient to prove
thus for b1, b2 being Element of the carrier of a1
st b1 <> b2 & b1 in Cl {b2}
holds not b2 in Cl {b1};
:: TSP_1:def 6
theorem
for b1 being non empty TopSpace-like TopStruct holds
b1 is discerning
iff
for b2, b3 being Element of the carrier of b1
st b2 <> b3 & b2 in Cl {b3}
holds not b3 in Cl {b2};
:: TSP_1:attrnot 6 => T_0TOPSP:attr 2
definition
let a1 be TopStruct;
attr a1 is T_0 means
for b1, b2 being Element of the carrier of a1
st b1 <> b2 & b1 in Cl {b2}
holds not Cl {b2} c= Cl {b1};
end;
:: TSP_1:dfs 7
definiens
let a1 be non empty TopSpace-like TopStruct;
To prove
a1 is discerning
it is sufficient to prove
thus for b1, b2 being Element of the carrier of a1
st b1 <> b2 & b1 in Cl {b2}
holds not Cl {b2} c= Cl {b1};
:: TSP_1:def 7
theorem
for b1 being non empty TopSpace-like TopStruct holds
b1 is discerning
iff
for b2, b3 being Element of the carrier of b1
st b2 <> b3 & b2 in Cl {b3}
holds not Cl {b3} c= Cl {b2};
:: TSP_1:condreg 8
registration
cluster non empty TopSpace-like almost_discrete discerning -> discrete (TopStruct);
end;
:: TSP_1:condreg 9
registration
cluster non empty TopSpace-like non discrete almost_discrete -> non discerning (TopStruct);
end;
:: TSP_1:condreg 10
registration
cluster non empty TopSpace-like non discrete discerning -> non almost_discrete (TopStruct);
end;
:: TSP_1:modenot 3
definition
mode Kolmogorov_space is non empty TopSpace-like discerning TopStruct;
end;
:: TSP_1:modenot 4
definition
mode non-Kolmogorov_space is non empty TopSpace-like non discerning TopStruct;
end;
:: TSP_1:exreg 3
registration
cluster non empty non trivial strict TopSpace-like discerning TopStruct;
end;
:: TSP_1:exreg 4
registration
cluster non empty non trivial strict TopSpace-like non discrete non discerning TopStruct;
end;
:: TSP_1:attrnot 7 => TSP_1:attr 1
definition
let a1 be TopStruct;
let a2 be Element of bool the carrier of a1;
attr a2 is T_0 means
for b1, b2 being Element of the carrier of a1
st b1 in a2 &
b2 in a2 &
b1 <> b2 &
(for b3 being Element of bool the carrier of a1
st b3 is open(a1) & b1 in b3
holds b2 in b3)
holds ex b3 being Element of bool the carrier of a1 st
b3 is open(a1) & not b1 in b3 & b2 in b3;
end;
:: TSP_1:dfs 8
definiens
let a1 be TopStruct;
let a2 be Element of bool the carrier of a1;
To prove
a2 is T_0
it is sufficient to prove
thus for b1, b2 being Element of the carrier of a1
st b1 in a2 &
b2 in a2 &
b1 <> b2 &
(for b3 being Element of bool the carrier of a1
st b3 is open(a1) & b1 in b3
holds b2 in b3)
holds ex b3 being Element of bool the carrier of a1 st
b3 is open(a1) & not b1 in b3 & b2 in b3;
:: TSP_1:def 8
theorem
for b1 being TopStruct
for b2 being Element of bool the carrier of b1 holds
b2 is T_0(b1)
iff
for b3, b4 being Element of the carrier of b1
st b3 in b2 &
b4 in b2 &
b3 <> b4 &
(for b5 being Element of bool the carrier of b1
st b5 is open(b1) & b3 in b5
holds b4 in b5)
holds ex b5 being Element of bool the carrier of b1 st
b5 is open(b1) & not b3 in b5 & b4 in b5;
:: TSP_1:attrnot 8 => TSP_1:attr 1
definition
let a1 be TopStruct;
let a2 be Element of bool the carrier of a1;
attr a2 is T_0 means
for b1, b2 being Element of the carrier of a1
st b1 in a2 &
b2 in a2 &
b1 <> b2 &
(for b3 being Element of bool the carrier of a1
st b3 is closed(a1) & b1 in b3
holds b2 in b3)
holds ex b3 being Element of bool the carrier of a1 st
b3 is closed(a1) & not b1 in b3 & b2 in b3;
end;
:: TSP_1:dfs 9
definiens
let a1 be non empty TopStruct;
let a2 be Element of bool the carrier of a1;
To prove
a2 is T_0
it is sufficient to prove
thus for b1, b2 being Element of the carrier of a1
st b1 in a2 &
b2 in a2 &
b1 <> b2 &
(for b3 being Element of bool the carrier of a1
st b3 is closed(a1) & b1 in b3
holds b2 in b3)
holds ex b3 being Element of bool the carrier of a1 st
b3 is closed(a1) & not b1 in b3 & b2 in b3;
:: TSP_1:def 9
theorem
for b1 being non empty TopStruct
for b2 being Element of bool the carrier of b1 holds
b2 is T_0(b1)
iff
for b3, b4 being Element of the carrier of b1
st b3 in b2 &
b4 in b2 &
b3 <> b4 &
(for b5 being Element of bool the carrier of b1
st b5 is closed(b1) & b3 in b5
holds b4 in b5)
holds ex b5 being Element of bool the carrier of b1 st
b5 is closed(b1) & not b3 in b5 & b4 in b5;
:: TSP_1:th 5
theorem
for b1, b2 being TopStruct
for b3 being Element of bool the carrier of b1
for b4 being Element of bool the carrier of b2
st TopStruct(#the carrier of b1,the topology of b1#) = TopStruct(#the carrier of b2,the topology of b2#) &
b3 = b4 &
b3 is T_0(b1)
holds b4 is T_0(b2);
:: TSP_1:th 6
theorem
for b1 being non empty TopStruct
for b2 being Element of bool the carrier of b1
st b2 = the carrier of b1
holds b2 is T_0(b1)
iff
b1 is discerning;
:: TSP_1:th 7
theorem
for b1 being non empty TopStruct
for b2, b3 being Element of bool the carrier of b1
st b3 c= b2 & b2 is T_0(b1)
holds b3 is T_0(b1);
:: TSP_1:th 8
theorem
for b1 being non empty TopStruct
for b2, b3 being Element of bool the carrier of b1
st (b2 is T_0(b1) or b3 is T_0(b1))
holds b2 /\ b3 is T_0(b1);
:: TSP_1:th 9
theorem
for b1 being non empty TopStruct
for b2, b3 being Element of bool the carrier of b1
st (b2 is open(b1) or b3 is open(b1)) & b2 is T_0(b1) & b3 is T_0(b1)
holds b2 \/ b3 is T_0(b1);
:: TSP_1:th 10
theorem
for b1 being non empty TopStruct
for b2, b3 being Element of bool the carrier of b1
st (b2 is closed(b1) or b3 is closed(b1)) & b2 is T_0(b1) & b3 is T_0(b1)
holds b2 \/ b3 is T_0(b1);
:: TSP_1:th 11
theorem
for b1 being non empty TopStruct
for b2 being Element of bool the carrier of b1
st b2 is discrete(b1)
holds b2 is T_0(b1);
:: TSP_1:th 12
theorem
for b1 being non empty TopStruct
for b2 being non empty Element of bool the carrier of b1
st b2 is anti-discrete(b1) & b2 is not trivial
holds b2 is not T_0(b1);
:: TSP_1:attrnot 9 => TSP_1:attr 1
definition
let a1 be TopStruct;
let a2 be Element of bool the carrier of a1;
attr a2 is T_0 means
for b1, b2 being Element of the carrier of a1
st b1 in a2 & b2 in a2 & b1 <> b2
holds Cl {b1} <> Cl {b2};
end;
:: TSP_1:dfs 10
definiens
let a1 be non empty TopSpace-like TopStruct;
let a2 be Element of bool the carrier of a1;
To prove
a2 is T_0
it is sufficient to prove
thus for b1, b2 being Element of the carrier of a1
st b1 in a2 & b2 in a2 & b1 <> b2
holds Cl {b1} <> Cl {b2};
:: TSP_1:def 10
theorem
for b1 being non empty TopSpace-like TopStruct
for b2 being Element of bool the carrier of b1 holds
b2 is T_0(b1)
iff
for b3, b4 being Element of the carrier of b1
st b3 in b2 & b4 in b2 & b3 <> b4
holds Cl {b3} <> Cl {b4};
:: TSP_1:attrnot 10 => TSP_1:attr 1
definition
let a1 be TopStruct;
let a2 be Element of bool the carrier of a1;
attr a2 is T_0 means
for b1, b2 being Element of the carrier of a1
st b1 in a2 & b2 in a2 & b1 <> b2 & b1 in Cl {b2}
holds not b2 in Cl {b1};
end;
:: TSP_1:dfs 11
definiens
let a1 be non empty TopSpace-like TopStruct;
let a2 be Element of bool the carrier of a1;
To prove
a2 is T_0
it is sufficient to prove
thus for b1, b2 being Element of the carrier of a1
st b1 in a2 & b2 in a2 & b1 <> b2 & b1 in Cl {b2}
holds not b2 in Cl {b1};
:: TSP_1:def 11
theorem
for b1 being non empty TopSpace-like TopStruct
for b2 being Element of bool the carrier of b1 holds
b2 is T_0(b1)
iff
for b3, b4 being Element of the carrier of b1
st b3 in b2 & b4 in b2 & b3 <> b4 & b3 in Cl {b4}
holds not b4 in Cl {b3};
:: TSP_1:attrnot 11 => TSP_1:attr 1
definition
let a1 be TopStruct;
let a2 be Element of bool the carrier of a1;
attr a2 is T_0 means
for b1, b2 being Element of the carrier of a1
st b1 in a2 & b2 in a2 & b1 <> b2 & b1 in Cl {b2}
holds not Cl {b2} c= Cl {b1};
end;
:: TSP_1:dfs 12
definiens
let a1 be non empty TopSpace-like TopStruct;
let a2 be Element of bool the carrier of a1;
To prove
a2 is T_0
it is sufficient to prove
thus for b1, b2 being Element of the carrier of a1
st b1 in a2 & b2 in a2 & b1 <> b2 & b1 in Cl {b2}
holds not Cl {b2} c= Cl {b1};
:: TSP_1:def 12
theorem
for b1 being non empty TopSpace-like TopStruct
for b2 being Element of bool the carrier of b1 holds
b2 is T_0(b1)
iff
for b3, b4 being Element of the carrier of b1
st b3 in b2 & b4 in b2 & b3 <> b4 & b3 in Cl {b4}
holds not Cl {b4} c= Cl {b3};
:: TSP_1:th 13
theorem
for b1 being non empty TopSpace-like TopStruct
for b2 being empty Element of bool the carrier of b1 holds
b2 is T_0(b1);
:: TSP_1:th 14
theorem
for b1 being non empty TopSpace-like TopStruct
for b2 being Element of the carrier of b1 holds
{b2} is T_0(b1);
:: TSP_1:exreg 5
registration
let a1 be non empty TopStruct;
cluster non empty strict discerning SubSpace of a1;
end;
:: TSP_1:attrnot 12 => TSP_1:attr 2
definition
let a1 be TopStruct;
let a2 be SubSpace of a1;
redefine attr a2 is T_0 means
(a2 is not empty) implies for b1, b2 being Element of the carrier of a1
st b1 is Element of the carrier of a2 &
b2 is Element of the carrier of a2 &
b1 <> b2 &
(for b3 being Element of bool the carrier of a1
st b3 is open(a1) & b1 in b3
holds b2 in b3)
holds ex b3 being Element of bool the carrier of a1 st
b3 is open(a1) & not b1 in b3 & b2 in b3;
end;
:: TSP_1:dfs 13
definiens
let a1 be TopStruct;
let a2 be SubSpace of a1;
To prove
a1 is discerning
it is sufficient to prove
thus (a2 is not empty) implies for b1, b2 being Element of the carrier of a1
st b1 is Element of the carrier of a2 &
b2 is Element of the carrier of a2 &
b1 <> b2 &
(for b3 being Element of bool the carrier of a1
st b3 is open(a1) & b1 in b3
holds b2 in b3)
holds ex b3 being Element of bool the carrier of a1 st
b3 is open(a1) & not b1 in b3 & b2 in b3;
:: TSP_1:def 13
theorem
for b1 being TopStruct
for b2 being SubSpace of b1 holds
b2 is discerning
iff
(b2 is empty or for b3, b4 being Element of the carrier of b1
st b3 is Element of the carrier of b2 &
b4 is Element of the carrier of b2 &
b3 <> b4 &
(for b5 being Element of bool the carrier of b1
st b5 is open(b1) & b3 in b5
holds b4 in b5)
holds ex b5 being Element of bool the carrier of b1 st
b5 is open(b1) & not b3 in b5 & b4 in b5);
:: TSP_1:attrnot 13 => TSP_1:attr 3
definition
let a1 be TopStruct;
let a2 be SubSpace of a1;
redefine attr a2 is T_0 means
(a2 is not empty) implies for b1, b2 being Element of the carrier of a1
st b1 is Element of the carrier of a2 &
b2 is Element of the carrier of a2 &
b1 <> b2 &
(for b3 being Element of bool the carrier of a1
st b3 is closed(a1) & b1 in b3
holds b2 in b3)
holds ex b3 being Element of bool the carrier of a1 st
b3 is closed(a1) & not b1 in b3 & b2 in b3;
end;
:: TSP_1:dfs 14
definiens
let a1 be TopStruct;
let a2 be SubSpace of a1;
To prove
a1 is discerning
it is sufficient to prove
thus (a2 is not empty) implies for b1, b2 being Element of the carrier of a1
st b1 is Element of the carrier of a2 &
b2 is Element of the carrier of a2 &
b1 <> b2 &
(for b3 being Element of bool the carrier of a1
st b3 is closed(a1) & b1 in b3
holds b2 in b3)
holds ex b3 being Element of bool the carrier of a1 st
b3 is closed(a1) & not b1 in b3 & b2 in b3;
:: TSP_1:def 14
theorem
for b1 being TopStruct
for b2 being SubSpace of b1 holds
b2 is discerning
iff
(b2 is empty or for b3, b4 being Element of the carrier of b1
st b3 is Element of the carrier of b2 &
b4 is Element of the carrier of b2 &
b3 <> b4 &
(for b5 being Element of bool the carrier of b1
st b5 is closed(b1) & b3 in b5
holds b4 in b5)
holds ex b5 being Element of bool the carrier of b1 st
b5 is closed(b1) & not b3 in b5 & b4 in b5);
:: TSP_1:th 15
theorem
for b1 being non empty TopStruct
for b2 being non empty SubSpace of b1
for b3 being Element of bool the carrier of b1
st b3 = the carrier of b2
holds b3 is T_0(b1)
iff
b2 is discerning;
:: TSP_1:th 16
theorem
for b1 being non empty TopStruct
for b2 being non empty SubSpace of b1
for b3 being non empty discerning SubSpace of b1
st b2 is SubSpace of b3
holds b2 is discerning;
:: TSP_1:th 17
theorem
for b1 being non empty TopSpace-like TopStruct
for b2 being non empty discerning SubSpace of b1
for b3 being non empty SubSpace of b1
st b2 meets b3
holds b2 meet b3 is discerning;
:: TSP_1:th 18
theorem
for b1 being non empty TopSpace-like TopStruct
for b2, b3 being non empty discerning SubSpace of b1
st (b2 is open(b1) or b3 is open(b1))
holds b2 union b3 is discerning;
:: TSP_1:th 19
theorem
for b1 being non empty TopSpace-like TopStruct
for b2, b3 being non empty discerning SubSpace of b1
st (b2 is closed(b1) or b3 is closed(b1))
holds b2 union b3 is discerning;
:: TSP_1:modenot 5
definition
let a1 be non empty TopSpace-like TopStruct;
mode Kolmogorov_subspace of a1 is non empty discerning SubSpace of a1;
end;
:: TSP_1:th 20
theorem
for b1 being non empty TopSpace-like TopStruct
for b2 being non empty Element of bool the carrier of b1
st b2 is T_0(b1)
holds ex b3 being non empty strict discerning SubSpace of b1 st
b2 = the carrier of b3;
:: TSP_1:exreg 6
registration
let a1 be non empty non trivial TopSpace-like TopStruct;
cluster non empty strict proper discerning SubSpace of a1;
end;
:: TSP_1:condreg 11
registration
let a1 be non empty TopSpace-like discerning TopStruct;
cluster non empty -> discerning (SubSpace of a1);
end;
:: TSP_1:condreg 12
registration
let a1 be non empty TopSpace-like non discerning TopStruct;
cluster non empty non proper -> non discerning (SubSpace of a1);
end;
:: TSP_1:condreg 13
registration
let a1 be non empty TopSpace-like non discerning TopStruct;
cluster non empty discerning -> proper (SubSpace of a1);
end;
:: TSP_1:exreg 7
registration
let a1 be non empty TopSpace-like non discerning TopStruct;
cluster strict non discerning SubSpace of a1;
end;
:: TSP_1:modenot 6
definition
let a1 be non empty TopSpace-like non discerning TopStruct;
mode non-Kolmogorov_subspace of a1 is non discerning SubSpace of a1;
end;
:: TSP_1:th 21
theorem
for b1 being non empty TopSpace-like non discerning TopStruct
for b2 being Element of bool the carrier of b1
st b2 is not T_0(b1)
holds ex b3 being strict non discerning SubSpace of b1 st
b2 = the carrier of b3;