anonymous
  • anonymous
A topological space X satifies the first separation axiom T_{1 }if each one of any two points of X has a nbd that does not containthe other point. Thus,X is called a T_{1} - space otherwise known as ______________ Metric Space Hausdorff Space Frechet Space Topological Space
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
i think frechet space
anonymous
  • anonymous
@zzr0ck3r
zzr0ck3r
  • zzr0ck3r
correct

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anonymous
  • anonymous
One of the conditions below is not a condition fot a topological space X to satisfy the first separation axiom If and only if all one point set in X is closed If and only if all one point set in X is open If and only if every finite set in is closed If and only if for x, y \epsilon X x\epsilon U_{x}and y\epsilon U_{y}
anonymous
  • anonymous
@zzr0ck3r
zzr0ck3r
  • zzr0ck3r
chill, reading
zzr0ck3r
  • zzr0ck3r
1) and 3) are true, I cant tell what you mean by the last line
anonymous
  • anonymous
\[Iff for x, y \epsilon X x\epsilon U_{x}and y\epsilon U_{y} \] the last statement
zzr0ck3r
  • zzr0ck3r
\[\forall x,y\in X \ x\in U_x\text{ and } y\in U_y\] What are \(U_x, U_y\) ?
zzr0ck3r
  • zzr0ck3r
It has to be the open one.
zzr0ck3r
  • zzr0ck3r
If every one point set was open, then we would just get the discrete topology every time, and thus the T_1 definition would be redundant.
zzr0ck3r
  • zzr0ck3r
What open sets?
zzr0ck3r
  • zzr0ck3r
\[\epsilon \neq \in\]
zzr0ck3r
  • zzr0ck3r
Don't une `\epsilon` to mean "in" :)
anonymous
  • anonymous
Ux and Uy are open sets not given the open sets sir . ok. what about option 2. is it right?
anonymous
  • anonymous
thanks for tell me that sir
zzr0ck3r
  • zzr0ck3r
2) is not true
zzr0ck3r
  • zzr0ck3r
unless it is finite.
zzr0ck3r
  • zzr0ck3r
lol
anonymous
  • anonymous
Let X and Y be topological spaces; let f: X\rightarrow Y be a bijection. If both f and its inverse f^{-1} are continuous, then f is called _____________________ Epimorphism Homomorphism Homeomorphism Endomorpism
zzr0ck3r
  • zzr0ck3r
homeomorphism
zzr0ck3r
  • zzr0ck3r
homomorphism is a group concept
anonymous
  • anonymous
sorry C
zzr0ck3r
  • zzr0ck3r
correct
anonymous
  • anonymous
A topological space is called a Hausdorff space, if for each x, y of distinct points of X, there exists nbds U_{x} and U_{y} of x and y respectively that are disjoint. This implies X is Hausdorff with these properties except one.
anonymous
  • anonymous
sorry wrong option sir
anonymous
  • anonymous
a \[If \forall x, y \epsilon\mathbb X; x \neg y \]
anonymous
  • anonymous
b \[There \exists U_x \epsilon N(x). U_y \epsilon N(y) \]
anonymous
  • anonymous
c \[ \forall x, y \epsilon X, x \bigcap y = 0 \]
zzr0ck3r
  • zzr0ck3r
what is \(\neg\)?
anonymous
  • anonymous
i think they want to confuse me with that or its an error
anonymous
  • anonymous
\[U_x \bigcap U-y = \phi \]
anonymous
  • anonymous
to me C is true
anonymous
  • anonymous
B is also true
anonymous
  • anonymous
but a and d, i am confuse
zzr0ck3r
  • zzr0ck3r
these do not make sense The first one: What the hell is \(\neg\) The second one: \(U_x\in N(x)\) makes no sense, Sets are not elements in the neighborhood, they are subsets if anything. The third one: Are we to assume \(x\ne y\)? Because if they are equal then this is of course true. The last one should say \(U_x\cap U_y=\emptyset\) and this is true.
anonymous
  • anonymous
ok sir, what if option A was \[If \forall x, y \epsilon\mathbb X; x \rightarrow y \] is it correct?
zzr0ck3r
  • zzr0ck3r
can you take a screen shot?
zzr0ck3r
  • zzr0ck3r
you latex work needs some love :)
anonymous
  • anonymous
lol
anonymous
  • anonymous
i will but its not always clear
zzr0ck3r
  • zzr0ck3r
here is one more thing for latest. Dont use words in the math part Like if the following: If `\(x\in X\)` then we have `\(\forall x \)`
zzr0ck3r
  • zzr0ck3r
keep the text outside of the inline math. or else it ill look like this \(if x\in X then \forall x\) Which was given by the following code `\(if x\in X then \forall x\)`
zzr0ck3r
  • zzr0ck3r
or put the text in `\text{stuff here}` Example \(\text{if }x\in X\text{ then } \forall x\)
zzr0ck3r
  • zzr0ck3r
ok close this and ask a new one if you have something else.
zzr0ck3r
  • zzr0ck3r
tag me if you need me...
anonymous
  • anonymous
There exists U_x \epsilon N(x). U_y \epsilon N(y) here is how option B is
anonymous
  • anonymous
@zzr0ck3r
anonymous
  • anonymous
is epsilon not same as in?
zzr0ck3r
  • zzr0ck3r
nope. epsilon is a greek letter, in is just means ... in :)
zzr0ck3r
  • zzr0ck3r
That option makes no sense...
anonymous
  • anonymous
it din't show fully but thats how it is.ok. thank you sir
zzr0ck3r
  • zzr0ck3r
or it is trivially true. can you say in normal words what that option says?
zzr0ck3r
  • zzr0ck3r
Like, generally \(U_X\) means the following: An open set \(U\) that contains \(x\). But this is always true in a topological space..
anonymous
  • anonymous
hmmm
zzr0ck3r
  • zzr0ck3r
Are you on windows?
anonymous
  • anonymous
but, so, that option is wrong since a set containing X can not be in the neighborhood of x
anonymous
  • anonymous
no sir.
zzr0ck3r
  • zzr0ck3r
Why cant a set containing \(x\) not be in \(N(x)\)?
zzr0ck3r
  • zzr0ck3r
\(N(x)\subseteq N(x)\). Unless you mean \(\in\) in which case it is weird....
anonymous
  • anonymous
N(x) means elements around X and not sets around X. i might be wrong
anonymous
  • anonymous
ok. can we try another?
zzr0ck3r
  • zzr0ck3r
yep, yes then, if you mean \(\in\) and not \(\epsilon\), then yes it does not make sense. But it actually could This will be a little weird but you elements could actually be sets. \(\{x, \{x\}\}\) is a thing. and \(\{x\}\in \{x,\{x\}\}\) But lets move on.
zzr0ck3r
  • zzr0ck3r
close this please, it takes to long to scroll on this pc

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