Question

anyone who've taken real analysis? I have question to ask..

Answer 1

Ask it.

Answer 2

is my proof rigorous enough?
Let f:X->R, X be subset of R, E be subset of X, x0 be adherent point of E

Answer 3

then f converges to L at x0 in E if and only if for every sequence (a_n)_{n=0}^{infinity} where a_n is an element of E for all n>=0, and converges to x0, (f(a_n))_{n=0}^{infinity} converges to L

Answer 4

Proof:
by definition, for every T>0, f is always T close to L near x0
by definition again, for every T>0, there exists Y>0 such that |L-f(x)|<=T for all x which is an element of the set {x' element of E:|x'-x0|<Y}.
now for any Y>0, x0 will always be in such set since |x0-x0|=0, so |x0-x0|<Y for any Y.
hence |L-f(x0)|<=Y for all Y>0. now since x0 is a constant number, then f(x0) is constant by a previous definition and results. by a prior proposition, this means f(x0)=L.
Now since x0 is an adherent point of E, then by a prior lemma, there exists sequence (a_n)_{n=0}^{infinity} which converges to x0.

Answer 5

by a previously proven fact again, for every U>0 there exists N>=0 such that |x0-a_n|<=U for all n>=N. This means by a previously proven result that x0=a_n for all n>=N. which implies that f(x0)=f(a_n) for all n>=N, since x0=a_n for all n>=N. since N>=0, that means however that that for any I>0, |f(x0)-f(a_n)|<=I, for all n>=N. This means that (f(x_n))_{n=0}^{infinity} converges to f(x0) by definition. since f(x0)=L, the sequence (f(x_n))_{n=0}^{infinity} converges to L.

Answer 6

\(
f: X\to R \\
E\subset X\\
x_0\in \bar E\\
L\in R\\
\text {restrict f to } \bar E\\
f:\bar E \to R\\
\lim_{x\to x_0} f(x)=L \iff \lim_{n\to \infty}f(x_n) =L
\)

for every sequence \( (x_n)\) in \(E\) converging to \(x_0\).


Let is proves the \( \implies \) part

We have
\(\lim_{x\to x_0} f(x)=L
\)

and a sequence \( (x_n)\) in \(E\) converging to \(x_0\).

For \( \epsilon > 0 \), there is a \( \delta>0\) so that if
\[
x\in E,\, |x-x_0| \le \delta \implies |f(x) - L|<\epsilon\]
Since \( (x_n)\) in \(E\) converging to \(x_0\), then
\( \exists\, N > 0,\) such that if
\[
n\ge N,\, |x_n -x_0|\leq \delta
\implies |f(x_n) -L| <\epsilon \]
This proves this part.

Let us prove the other implication.
We have to show that if for every sequence \( (x_n)\) in \(E\) converging to \(x_0\), \(\lim_{n\to \infty}f(x_n) =L\), then
\(\lim_{x\to x_0} f(x)=L
\)
Let us prove this part by contradiction.
There is \( \epsilon >0 \)
For every \(n >0\, \exists \, x_n\in E \, |x_n -x_0|<\frac 1 n \) and
\[ |f(x_n) - L|>\epsilon\]
That is a contradiction, since the sequence \( x_n\) converges to \( x_0\) but the the sequence \( f(x_n)\) does not converge to \(L\)