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24m
revised To write the sum of the given series
edited tags
1h
comment open set in subspace
The OP's conclusion is partly correct, but this is merely a fortunate accident, since the reasoning is entirely wrong.
1h
comment a multiset where two consecutive numbers must be dissimilar
@anonymous: You're welcome.
1h
awarded  Enlightened
1h
comment open set in subspace
I'm afraid not: no singleton $\{x\}$ is open in $\Bbb R$. A non-empty subset of $\Bbb R$ is open if and only if it is a union of open intervals.
2h
answered a multiset where two consecutive numbers must be dissimilar
2h
answered a simple question about topological space
3h
comment Every point in a Tychonoff Space is contained in a compact set
You’re welcome!
3h
answered Difference between “distribution” & “arrangement”.
3h
comment Every point in a Tychonoff Space is contained in a compact set
In any space $\{x\}$ is compact for each point $x$ in the space. Are you sure that you're asking the right question?
3h
comment generalize the question every every intersection of nested sequence of compact non-empty sets is compact and non-empty
@F.K: Glad to hear it. You're welcome!
3h
comment Prove that the following is a constant function
@CiaPan: The whole point was not to say it explicitly, but rather to give, as the OP said, a strong hint -- something that would point the way reasonably clearly but still require some thought.
3h
awarded  Nice Answer
4h
comment What is the pattern of this sequence of numbers?
In view of the OP's clarification 14 hours ago, putting the question on hold 11 hours ago was inexcusable.
6h
reviewed Edit Number of onto functions
6h
revised Number of onto functions
Removed typo.
7h
answered Number of Hausdorff topologies on a set with $100$ elements.
7h
comment Questions concerning elements in $F = \big\{f: \{1, 2, 3\} \to \{1, 2, 3, 4, 5\}\big\}$.
You’re welcome!
7h
comment Questions concerning elements in $F = \big\{f: \{1, 2, 3\} \to \{1, 2, 3, 4, 5\}\big\}$.
Yes, it is. You could also have done the second problem in one step: there are $2$ choices for $f(1)$, and no matter how you make that choice, there are $4$ for $f(2)$ and then $3$ for $f(3)$, for a total of $2\cdot4\cdot3=24$.
7h
awarded  analysis