let X be Subset of REAL; ( ( for r being Element of REAL st r in X holds
ex N being Neighbourhood of r st N c= X ) implies X is open )
assume that
A1:
for r being Element of REAL st r in X holds
ex N being Neighbourhood of r st N c= X
and
A2:
not X is open
; contradiction
not X ` is closed
by A2;
then consider s1 being Real_Sequence such that
A3:
rng s1 c= X `
and
A4:
s1 is convergent
and
A5:
not lim s1 in X `
;
reconsider ls = lim s1 as Element of REAL by XREAL_0:def 1;
consider N being Neighbourhood of ls such that
A6:
N c= X
by A1, A5, SUBSET_1:29;
consider g being Real such that
A7:
0 < g
and
A8:
].((lim s1) - g),((lim s1) + g).[ = N
by Def6;
consider n being Nat such that
A9:
for m being Nat st n <= m holds
|.((s1 . m) - (lim s1)).| < g
by A4, A7, SEQ_2:def 7;
n in NAT
by ORDINAL1:def 12;
then
n in dom s1
by FUNCT_2:def 1;
then A10:
s1 . n in rng s1
by FUNCT_1:def 3;
A11:
|.((s1 . n) - (lim s1)).| < g
by A9;
then
(s1 . n) - (lim s1) < g
by SEQ_2:1;
then A12:
s1 . n < (lim s1) + g
by XREAL_1:19;
- g < (s1 . n) - (lim s1)
by A11, SEQ_2:1;
then
(lim s1) + (- g) < (lim s1) + ((s1 . n) - (lim s1))
by XREAL_1:6;
then
s1 . n in { s where s is Real : ( (lim s1) - g < s & s < (lim s1) + g ) }
by A12;
hence
contradiction
by A3, A6, A8, A10, XBOOLE_0:def 5; verum