help

- anonymous

help

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- schrodinger

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

1 Every complete metric space is a ___________
Baire space
Blank space
Dense space
Cardinal space

- anonymous

Baire space :)

- anonymous

@Michele_Laino

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

i am here sir

- Michele_Laino

ok!
a complete space is a space in which every Cauchy sequence converges in it

- anonymous

so, which means that Every complete metric space is a ___________
Baire space

- anonymous

don't get angry at me but what is a Cauchy sequence ?

- Michele_Laino

a sequence is said a Cauchy sequence, if given \epsilon>0, there exists a natual number, say N, such thatfor each n, m two natural numbers, greater or equal to N, the subsequent condition holds:
\[\Large d_X\left( {{x_n},{x_m}} \right) < \varepsilon \]
where
\[\Large {d_X}\]
is the distance of your metric space X

- Michele_Laino

now, every converget sequence is also a Cauchy sequence

- anonymous

ok

- Michele_Laino

furthermore, if each Cauchy sequence converges to an element x of X, since X is complete

- Michele_Laino

oops..I have made a typo, here is the right statement:
"each Cauchy sequence converges to an element x of X, since X is complete"

- anonymous

ok. thanks for that . now can we answer the questions i asked together sir?

- anonymous

i noticed it was a typo

- Michele_Laino

since each convergent sequence is a Cauchy sequence, and each Cauchy sequence converges to an element of the space X it self, then X contains all its limits point, in other word, we have:
\[\Large \overline X = X\]
so X is a dense space

- anonymous

which means that a complete metric space is a dense space

- Michele_Laino

yes!

- anonymous

thanks so much sir

- Michele_Laino

ok! let's go to the next question, please

- anonymous

2 Let
(X,τ)
be a topological space. If
X
is second countable, then
X
is ____________countable
Third
Second
Fourth
First

- anonymous

i think first

- Michele_Laino

If a topological space is second countable, then it is also first countable, sincce the second countability implies the first countability

- Michele_Laino

since*

- anonymous

ok.

- anonymous

3 If
xϵA¯
, then there exists a sequence
(xn)
of A such that
xn→x
is only true if X is a(an) ___________________
Countable
Metrizable
Hausdorff
Separation

- anonymous

i think B

- Michele_Laino

yes! I think so, since in order to speak about limit, we need of a topology, which can be induced by the metric of the space, so we need of a metrizable space

- anonymous

thanks.

- anonymous

7 Let
X=(a,b,c,d,e)andτ=(X,ϕ,[a],[c,d],[a,c,d],[b,c,d,e]).LetA=[a,c]]
, then set
A’ of limit points of A
is given by
A′=(b,c,e)
A′=(b,d,e)
A′=(b,e)
A′=X

- anonymous

please , i don't know things on that

- anonymous

i think (b,d,e)

- Michele_Laino

x=a can not be a limit point

- anonymous

so, whatt should be the correct one

- Michele_Laino

yes! I think so, it is {b,d,e}

- anonymous

please why is it (b,d,e)

- anonymous

please explain sir

- Michele_Laino

since each neighborhood around x=b, d, e contains points of A other than b, d, e

- anonymous

ok sir

- anonymous

8 Let
R
, the real line be endowned with the discrete topology. Which of the following subsets of
R
is dense in
R
Q
Ritself
Qc
All singletons

- Michele_Laino

here it is Q is dense in R, since we can show that between two real numbers, exists a rational number

- anonymous

i really need your help in this topology. i wish we can make out study time

- anonymous

9 Let
A=(0,1]⋃2
be a subset of
R
. Then the isolated points of
AinR
are
0 and1
0 and 2
1 and 2
[2]

- Michele_Laino

x=0, 1 can not be isolated points

- anonymous

is it 0 and 2

- anonymous

and explain

- Michele_Laino

no, x=0, is a limit point of A

- anonymous

so which?

- Michele_Laino

[2] is a closed set in R, nevertheless it is an open subset of A, since it is given by the intersection between A and (-1, 4), so
I think [2]

- anonymous

hmm. so which are the limit points?

- Michele_Laino

I think the answer is [2]

- Michele_Laino

since I can find at least one neighborhood around x=2, such that it contains only x=2

- anonymous

ok thanks

- anonymous

10 For the set A in question above, which of the following are the limit points of
A
?
0 and1
0 and 2
1 and 2
2 only

- Michele_Laino

x=0, and x=1

- anonymous

sir can you teach me four things here??

- Michele_Laino

yes!

- anonymous

teach me the difference between usual real line and the standard real line

- Michele_Laino

In general with the real line we indicate the set of the real number without the points:
\[ \Large + \infty ,\quad - \infty \]
furthermore, when we add those points to the real line, we get the so called "expanded line"

- Michele_Laino

the so defined "expanded line" is again a totally ordered set

- anonymous

ok. please teach me the intersection and union of sets of real line. like

- anonymous

A=(0,1]⋃2

- Michele_Laino

they are defined as usually. Namely, the intersection of two sets, is set of all points which belong to both those sets.
Similarly for the union of two sets, which is the set of the points which belong to one set or to the other set or to both
In your case, I think better is:
A=(0,1]⋃[2]

- anonymous

ok

- anonymous

do they mean (0,2)?

- Michele_Laino

no, since the set (0,2) is:
|dw:1440443474555:dw|

- Michele_Laino

x=0, and x=2 are not included

- Michele_Laino

whereas the set A=(0,1] union [2], is:
|dw:1440443551194:dw|

- Michele_Laino

|dw:1440443643274:dw|

- anonymous

do what do they want us to do?

- anonymous

what do they want us to do?

- Michele_Laino

they are representation of the two sets above

- anonymous

ok

Looking for something else?

Not the answer you are looking for? Search for more explanations.