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

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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
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i think frechet space
correct

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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}
chill, reading
1) and 3) are true, I cant tell what you mean by the last line
\[Iff for x, y \epsilon X x\epsilon U_{x}and y\epsilon U_{y} \] the last statement
\[\forall x,y\in X \ x\in U_x\text{ and } y\in U_y\] What are \(U_x, U_y\) ?
It has to be the open one.
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.
What open sets?
\[\epsilon \neq \in\]
Don't une `\epsilon` to mean "in" :)
Ux and Uy are open sets not given the open sets sir . ok. what about option 2. is it right?
thanks for tell me that sir
2) is not true
unless it is finite.
lol
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
homeomorphism
homomorphism is a group concept
sorry C
correct
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.
sorry wrong option sir
a \[If \forall x, y \epsilon\mathbb X; x \neg y \]
b \[There \exists U_x \epsilon N(x). U_y \epsilon N(y) \]
c \[ \forall x, y \epsilon X, x \bigcap y = 0 \]
what is \(\neg\)?
i think they want to confuse me with that or its an error
\[U_x \bigcap U-y = \phi \]
to me C is true
B is also true
but a and d, i am confuse
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.
ok sir, what if option A was \[If \forall x, y \epsilon\mathbb X; x \rightarrow y \] is it correct?
can you take a screen shot?
you latex work needs some love :)
lol
i will but its not always clear
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 `\(\forall x \)`
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 `\(if x\in X then \forall x\)`
or put the text in `\text{stuff here}` Example \(\text{if }x\in X\text{ then } \forall x\)
ok close this and ask a new one if you have something else.
tag me if you need me...
There exists U_x \epsilon N(x). U_y \epsilon N(y) here is how option B is
is epsilon not same as in?
nope. epsilon is a greek letter, in is just means ... in :)
That option makes no sense...
it din't show fully but thats how it is.ok. thank you sir
or it is trivially true. can you say in normal words what that option says?
Like, generally \(U_X\) means the following: An open set \(U\) that contains \(x\). But this is always true in a topological space..
hmmm
Are you on windows?
but, so, that option is wrong since a set containing X can not be in the neighborhood of x
no sir.
Why cant a set containing \(x\) not be in \(N(x)\)?
\(N(x)\subseteq N(x)\). Unless you mean \(\in\) in which case it is weird....
N(x) means elements around X and not sets around X. i might be wrong
ok. can we try another?
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.
close this please, it takes to long to scroll on this pc

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