All Questions
10
questions
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124
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Converse of Bolzano Weierstrass Theorem
Bolzano Weierstrass Theorem (for sequences) states that Every bounded sequence has a limit point.
However the converse is not true i.e. there do exist unbounded sequence(s) having only one real limit ...
0
votes
3
answers
94
views
Is it true that there is a bijection $[0, 1) \to \mathbb{R}$?
Is there is a bijection from $[0,1)$ to $\mathbb{R}$?
I thought of an instance, $$\frac{\sqrt{x(1-x)}}{x-1}.$$
- 127
0
votes
1
answer
42
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Explain a confusing bound for the integral of a decreasing function.
I am reading a solution of an exercise. In the solution, it says the following:
Consider $g(x,t):=\frac{x}{(1+tx^{2})t^{\alpha}}$, where $x\in (0,\infty)$, $t=1,2,3,\cdots$ and $\alpha>\frac{1}{2}$...
- 5,930
4
votes
1
answer
177
views
Abstract concept tying real numbers to elementary functions?
Real numbers can be broken into two categories: rational vs. irrational. Irrational numbers can be approximated, but never fully represented by rational numbers.
Analytic functions have Taylor ...
- 247
3
votes
1
answer
177
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Suppose $A$ is a countable subset of $\mathbb{R}$. Show that $\mathbb{R} \sim \mathbb{R} \setminus A $.
Here is the question I am trying to answer: Let $\mathbb{R}$ denote the reals and suppose A is a countable subset of $\mathbb{R}$. Show that $\mathbb{R} \sim \mathbb{R} \setminus A $.
My attempt:
...
- 159
1
vote
2
answers
102
views
Why the following sequence of function does not converge uniformly at $[0, \infty)$
Why the following sequence of function does not converge uniformly at $[0, \infty)$ but converge uniformly for some $a>0, [a,\infty)$
$$f_n(x) := n^2x^2e^{-nx}$$
So I know the limit function $f$ ...
- 2,157
0
votes
0
answers
60
views
Different Alternate Representations of Functions
Could someone please point out a source with detailed steps / other pointers for different alternate representations of functions?
For example, I know of two such ways
1) Taylor Series Expansion
2)...
- 800
4
votes
1
answer
282
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Is the range of an injective function dense somewhere?
Consider an injective function $\,\,f:[0,1]\rightarrow[0,1].\,$ Then is it true that there always exists some non-empty open subinterval of $[0,1]$, such that $f([0,1])$ is dense in that subinterval? ...
- 2,397
1
vote
1
answer
39
views
For $f(x, y) = x-y$, is $f(K \times K)$ closed if $K$ is closed?
$f: \mathbb{R}\times \mathbb{R} \rightarrow \mathbb{R}$ such that $f(x, y) = x-y$. For $K \subset \mathbb{R}$ closed is $f(K\times K)$ closed?
For the closed interval this is straight forwardly true ...
3
votes
2
answers
160
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Is the function $\,f(x, y) = x-y\,$ closed?
Is the function $\,\,f: \mathbb{R}\times \mathbb{R} \rightarrow \mathbb{R}$, defined as $$f(x,y)=x-y,$$ closed?