Mixing
Questions about the proof of the intermediate value theorem
$begingroup$
Let$ f:[a,b]rightarrow mathbb{R}$ be continious. Then for every $y_0in[min(f(a),f(b)), max (f(a),f(b))] $ there exists at least one $x_0in[a,b]$ with $f(x_0)=y_0$.
Proof: Without loss of generality let $f(a)leq f(b)$.
We define the set $ M$
$M:={xin[a,b]:f(x)geq y_0}=f^{-1}([y_0,+infty))subseteq [a,b].$
Then $bin M$, thus $M neq emptyset$. Let $x_0 := inf M in [a,b]$. There exists a sequence $(x_n)$ in $M$ with $x_nrightarrow x_0$. Because $f$ is continious, we have $f(x_n)rightarrow f(x_0)$, and because $f(x_n)geq y_0$ we also have $f(x_0)geq y_0$, in conclusion $x_0in M$, thus $x_0= min M$.
I have understood the rest of the proof but if somebody wants me to write it down I will do it.
For an infinum we need a lower bound what would be the lower bound in this case, and how do we know that the infinum sits in the intervall? Also I don't understand this part "...and because $f(x_n)geq y_0$ we also have $f(x_0)geq y_0$..." Every element of the sequence is $geq y_0$, why must the same also apply to the Limit?
real-analysis
edited Jan 11 at 13:20
Song
12.2k630
asked Jan 11 at 13:05
RM777RM777
40112
$endgroup$
add a comment |
$begingroup$
Let$ f:[a,b]rightarrow mathbb{R}$ be continious. Then for every $y_0in[min(f(a),f(b)), max (f(a),f(b))] $ there exists at least one $x_0in[a,b]$ with $f(x_0)=y_0$.
Proof: Without loss of generality let $f(a)leq f(b)$.
We define the set $ M$
$M:={xin[a,b]:f(x)geq y_0}=f^{-1}([y_0,+infty))subseteq [a,b].$
Then $bin M$, thus $M neq emptyset$. Let $x_0 := inf M in [a,b]$. There exists a sequence $(x_n)$ in $M$ with $x_nrightarrow x_0$. Because $f$ is continious, we have $f(x_n)rightarrow f(x_0)$, and because $f(x_n)geq y_0$ we also have $f(x_0)geq y_0$, in conclusion $x_0in M$, thus $x_0= min M$.
I have understood the rest of the proof but if somebody wants me to write it down I will do it.
For an infinum we need a lower bound what would be the lower bound in this case, and how do we know that the infinum sits in the intervall? Also I don't understand this part "...and because $f(x_n)geq y_0$ we also have $f(x_0)geq y_0$..." Every element of the sequence is $geq y_0$, why must the same also apply to the Limit?
real-analysis
edited Jan 11 at 13:20
Song
12.2k630
asked Jan 11 at 13:05
RM777RM777
40112
$endgroup$
add a comment |
$begingroup$
Let$ f:[a,b]rightarrow mathbb{R}$ be continious. Then for every $y_0in[min(f(a),f(b)), max (f(a),f(b))] $ there exists at least one $x_0in[a,b]$ with $f(x_0)=y_0$.
Proof: Without loss of generality let $f(a)leq f(b)$.
We define the set $ M$
$M:={xin[a,b]:f(x)geq y_0}=f^{-1}([y_0,+infty))subseteq [a,b].$
Then $bin M$, thus $M neq emptyset$. Let $x_0 := inf M in [a,b]$. There exists a sequence $(x_n)$ in $M$ with $x_nrightarrow x_0$. Because $f$ is continious, we have $f(x_n)rightarrow f(x_0)$, and because $f(x_n)geq y_0$ we also have $f(x_0)geq y_0$, in conclusion $x_0in M$, thus $x_0= min M$.
I have understood the rest of the proof but if somebody wants me to write it down I will do it.
For an infinum we need a lower bound what would be the lower bound in this case, and how do we know that the infinum sits in the intervall? Also I don't understand this part "...and because $f(x_n)geq y_0$ we also have $f(x_0)geq y_0$..." Every element of the sequence is $geq y_0$, why must the same also apply to the Limit?
real-analysis
edited Jan 11 at 13:20
Song
12.2k630
asked Jan 11 at 13:05
RM777RM777
40112
$endgroup$
Let$ f:[a,b]rightarrow mathbb{R}$ be continious. Then for every $y_0in[min(f(a),f(b)), max (f(a),f(b))] $ there exists at least one $x_0in[a,b]$ with $f(x_0)=y_0$.
Proof: Without loss of generality let $f(a)leq f(b)$.
We define the set $ M$
$M:={xin[a,b]:f(x)geq y_0}=f^{-1}([y_0,+infty))subseteq [a,b].$
Then $bin M$, thus $M neq emptyset$. Let $x_0 := inf M in [a,b]$. There exists a sequence $(x_n)$ in $M$ with $x_nrightarrow x_0$. Because $f$ is continious, we have $f(x_n)rightarrow f(x_0)$, and because $f(x_n)geq y_0$ we also have $f(x_0)geq y_0$, in conclusion $x_0in M$, thus $x_0= min M$.
I have understood the rest of the proof but if somebody wants me to write it down I will do it.
For an infinum we need a lower bound what would be the lower bound in this case, and how do we know that the infinum sits in the intervall? Also I don't understand this part "...and because $f(x_n)geq y_0$ we also have $f(x_0)geq y_0$..." Every element of the sequence is $geq y_0$, why must the same also apply to the Limit?
real-analysis
real-analysis
edited Jan 11 at 13:20
Song
12.2k630
asked Jan 11 at 13:05
RM777RM777
40112
edited Jan 11 at 13:20
Song
12.2k630
asked Jan 11 at 13:05
RM777RM777
40112
edited Jan 11 at 13:20
Song
12.2k630
edited Jan 11 at 13:20
Song
12.2k630
edited Jan 11 at 13:20
Song
12.2k630
12.2k630
asked Jan 11 at 13:05
RM777RM777
40112
asked Jan 11 at 13:05
RM777RM777
40112
asked Jan 11 at 13:05
RM777RM777
40112
40112
add a comment |
add a comment |
2 Answers
2
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oldest
votes
$begingroup$
- Since $M subset [a,b]$ it follows that $a$ is a lower bound.
- For the same reason you have $aleq inf M$.
- In general it holds $c leq y_n stackrel{ntoinfty}{rightarrow} y_0 Rightarrow c leq y_0$. You can see this quickly by contradiction:
Assume $y_0 < c$. Choose $epsilon > 0$ with $y_0 + epsilon < c stackrel{y_n stackrel{ntoinfty}{rightarrow} y_0}{Longrightarrow} c leq y_n < y_0 + epsilon < c$ for sufficient large $n$. Contradiction!
answered Jan 11 at 13:19
trancelocationtrancelocation
11k1723
$endgroup$
add a comment |
$begingroup$
The infimum of a set lies in the closure of the set (see here). So $inf M$ lies in $overline M subseteq overline {[a, b]} = [a, b]$. Basically, the infimum is in $[a, b]$ because it is a closed interval.
For your other question, yes: if all elements of a converging sequence are $geq y_0$, then the limit is also $geq y_0$. The same thing holds for "$leq$". You can prove it by contradiction: assume the limit is $lneqq y$. So you have $d := y_0-f(x_0) gneqq 0$. By definition of convergence, there exists $N$ sufficiently big such that $f(x_n)-f(x_0) leq frac d2$, therefore $y_0 - f(x_n) = (y_0 - f(x_0)) + (f(x_0)-f(x_n)) gneqq d - frac d2 gneqq 0$ for $n > N$ and this is a contradiction.
answered Jan 11 at 13:19
BermudesBermudes
18713
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2 Answers
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2 Answers
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$begingroup$
- Since $M subset [a,b]$ it follows that $a$ is a lower bound.
- For the same reason you have $aleq inf M$.
- In general it holds $c leq y_n stackrel{ntoinfty}{rightarrow} y_0 Rightarrow c leq y_0$. You can see this quickly by contradiction:
Assume $y_0 < c$. Choose $epsilon > 0$ with $y_0 + epsilon < c stackrel{y_n stackrel{ntoinfty}{rightarrow} y_0}{Longrightarrow} c leq y_n < y_0 + epsilon < c$ for sufficient large $n$. Contradiction!
answered Jan 11 at 13:19
trancelocationtrancelocation
11k1723
$endgroup$
add a comment |
$begingroup$
- Since $M subset [a,b]$ it follows that $a$ is a lower bound.
- For the same reason you have $aleq inf M$.
- In general it holds $c leq y_n stackrel{ntoinfty}{rightarrow} y_0 Rightarrow c leq y_0$. You can see this quickly by contradiction:
Assume $y_0 < c$. Choose $epsilon > 0$ with $y_0 + epsilon < c stackrel{y_n stackrel{ntoinfty}{rightarrow} y_0}{Longrightarrow} c leq y_n < y_0 + epsilon < c$ for sufficient large $n$. Contradiction!
answered Jan 11 at 13:19
trancelocationtrancelocation
11k1723
$endgroup$
add a comment |
$begingroup$
- Since $M subset [a,b]$ it follows that $a$ is a lower bound.
- For the same reason you have $aleq inf M$.
- In general it holds $c leq y_n stackrel{ntoinfty}{rightarrow} y_0 Rightarrow c leq y_0$. You can see this quickly by contradiction:
Assume $y_0 < c$. Choose $epsilon > 0$ with $y_0 + epsilon < c stackrel{y_n stackrel{ntoinfty}{rightarrow} y_0}{Longrightarrow} c leq y_n < y_0 + epsilon < c$ for sufficient large $n$. Contradiction!
answered Jan 11 at 13:19
trancelocationtrancelocation
11k1723
$endgroup$
- Since $M subset [a,b]$ it follows that $a$ is a lower bound.
- For the same reason you have $aleq inf M$.
- In general it holds $c leq y_n stackrel{ntoinfty}{rightarrow} y_0 Rightarrow c leq y_0$. You can see this quickly by contradiction:
Assume $y_0 < c$. Choose $epsilon > 0$ with $y_0 + epsilon < c stackrel{y_n stackrel{ntoinfty}{rightarrow} y_0}{Longrightarrow} c leq y_n < y_0 + epsilon < c$ for sufficient large $n$. Contradiction!
answered Jan 11 at 13:19
trancelocationtrancelocation
11k1723
answered Jan 11 at 13:19
trancelocationtrancelocation
11k1723
answered Jan 11 at 13:19
trancelocationtrancelocation
11k1723
answered Jan 11 at 13:19
trancelocationtrancelocation
11k1723
11k1723
add a comment |
add a comment |
$begingroup$
The infimum of a set lies in the closure of the set (see here). So $inf M$ lies in $overline M subseteq overline {[a, b]} = [a, b]$. Basically, the infimum is in $[a, b]$ because it is a closed interval.
For your other question, yes: if all elements of a converging sequence are $geq y_0$, then the limit is also $geq y_0$. The same thing holds for "$leq$". You can prove it by contradiction: assume the limit is $lneqq y$. So you have $d := y_0-f(x_0) gneqq 0$. By definition of convergence, there exists $N$ sufficiently big such that $f(x_n)-f(x_0) leq frac d2$, therefore $y_0 - f(x_n) = (y_0 - f(x_0)) + (f(x_0)-f(x_n)) gneqq d - frac d2 gneqq 0$ for $n > N$ and this is a contradiction.
answered Jan 11 at 13:19
BermudesBermudes
18713
$endgroup$
add a comment |
$begingroup$
The infimum of a set lies in the closure of the set (see here). So $inf M$ lies in $overline M subseteq overline {[a, b]} = [a, b]$. Basically, the infimum is in $[a, b]$ because it is a closed interval.
For your other question, yes: if all elements of a converging sequence are $geq y_0$, then the limit is also $geq y_0$. The same thing holds for "$leq$". You can prove it by contradiction: assume the limit is $lneqq y$. So you have $d := y_0-f(x_0) gneqq 0$. By definition of convergence, there exists $N$ sufficiently big such that $f(x_n)-f(x_0) leq frac d2$, therefore $y_0 - f(x_n) = (y_0 - f(x_0)) + (f(x_0)-f(x_n)) gneqq d - frac d2 gneqq 0$ for $n > N$ and this is a contradiction.
answered Jan 11 at 13:19
BermudesBermudes
18713
$endgroup$
add a comment |
$begingroup$
The infimum of a set lies in the closure of the set (see here). So $inf M$ lies in $overline M subseteq overline {[a, b]} = [a, b]$. Basically, the infimum is in $[a, b]$ because it is a closed interval.
For your other question, yes: if all elements of a converging sequence are $geq y_0$, then the limit is also $geq y_0$. The same thing holds for "$leq$". You can prove it by contradiction: assume the limit is $lneqq y$. So you have $d := y_0-f(x_0) gneqq 0$. By definition of convergence, there exists $N$ sufficiently big such that $f(x_n)-f(x_0) leq frac d2$, therefore $y_0 - f(x_n) = (y_0 - f(x_0)) + (f(x_0)-f(x_n)) gneqq d - frac d2 gneqq 0$ for $n > N$ and this is a contradiction.
answered Jan 11 at 13:19
BermudesBermudes
18713
$endgroup$
The infimum of a set lies in the closure of the set (see here). So $inf M$ lies in $overline M subseteq overline {[a, b]} = [a, b]$. Basically, the infimum is in $[a, b]$ because it is a closed interval.
For your other question, yes: if all elements of a converging sequence are $geq y_0$, then the limit is also $geq y_0$. The same thing holds for "$leq$". You can prove it by contradiction: assume the limit is $lneqq y$. So you have $d := y_0-f(x_0) gneqq 0$. By definition of convergence, there exists $N$ sufficiently big such that $f(x_n)-f(x_0) leq frac d2$, therefore $y_0 - f(x_n) = (y_0 - f(x_0)) + (f(x_0)-f(x_n)) gneqq d - frac d2 gneqq 0$ for $n > N$ and this is a contradiction.
answered Jan 11 at 13:19
BermudesBermudes
18713
edited Jan 11 at 13:47
answered Jan 11 at 13:19
BermudesBermudes
18713
answered Jan 11 at 13:19
BermudesBermudes
18713
answered Jan 11 at 13:19
BermudesBermudes
18713
18713
add a comment |
add a comment |
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