Prove or disaprove the statements

[mathmari]

Hey!

Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be continuous and $A,B\subseteq \mathbb{R}$. Prove or disprove the following:

1. $A$ closed $\Rightarrow$ $f(A)$ closed
2. $B$ closed $\Rightarrow$ $f^{-1}(B)$ closed
3. $A$ bounded $\Rightarrow$ $f(A)$ bounded
4. $B$ bounded $\Rightarrow$ $f^{-1}(B)$ bounded

I have done the following:

We have that $D$ is closed $\iff$ for each convergent sequence $(x_n)$ in $D$ it holds that $\displaystyle{\lim_{n\rightarrow \infty} x_n\in D}$.

We also have that $f: D\rightarrow \mathbb{R}$ is bounded $\iff$ $f(D)$ is bounded. This is equivalent to $\exists c\geq 0 \ \ \forall x\in D : |f(x)|\leq c$.

1. What can we do in this case? (Wondering)
   $$$$
2. Let $(x_n)$ be a convergent sequence in $f^{-1}(B)$. We want to show that $\displaystyle{\lim_{n\rightarrow \infty}x_n\in f^{-1}(B)}$.
   Since $f$ is continuous, we have that $\displaystyle{f\left (\lim_{n\rightarrow \infty} x_n\right )=\lim_{n\rightarrow \infty} f(x_n)}$.
   Since $(x_n)\in f^{-1}(B)$ and $f$ is continuous we have that $f(x_n)\in B$. Is this correct? (Wondering)
   Since $B$ is closed we have that $\displaystyle{\lim_{n\rightarrow \infty}f(x_n)\in B}$.
   Therefore, we have that $\displaystyle{f\left (\lim_{n\rightarrow \infty}x_n\right )\in B}$. From that it implies that $\displaystyle{\lim_{n\rightarrow \infty}x_n\in f^{-1}(B)}$. This means that $f^{-1}(B)$ is closed.
   Is this correct? (Wondering)
   $$$$
3. Since $A$ is bounded, we have that it is upper and under bounded. There are $x,y$ such that $$x\leq a \leq y , \ \ \text{ for all } a\in A$$ Since $f$ is continuous we have that $$f(x)\leq f(a) \leq f(y) , \ \ \forall a\in A$$
   This means that in $A$ the function $f$ is bounded.
   Therefore, $f(A)$ is bounded.
   Is this correct? (Wondering)
   $$$$
4. What can we do in this case? (Wondering)

 

[caffeinemachine]

Hey!

Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be continuous and $A,B\subseteq \mathbb{R}$. Prove or disprove the following:

1. $A$ closed $\Rightarrow$ $f(A)$ closed
2. $B$ closed $\Rightarrow$ $f^{-1}(B)$ closed
3. $A$ bounded $\Rightarrow$ $f(A)$ bounded
4. $B$ bounded $\Rightarrow$ $f^{-1}(B)$ bounded

I have done the following:

We have that $D$ is closed $\iff$ for each convergent sequence $(x_n)$ in $D$ it holds that $\displaystyle{\lim_{n\rightarrow \infty} x_n\in D}$.

We also have that $f: D\rightarrow \mathbb{R}$ is bounded $\iff$ $f(D)$ is bounded. This is equivalent to $\exists c\geq 0 \ \ \forall x\in D : |f(x)|\leq c$.

1. What can we do in this case? (Wondering)
   $$$$

Consider $f(x)=\tan^{-1}(x)$. Then $f(\mathbf R)=(-\pi/2, \pi/2)$. Thus $(1)$ is not true.

1. Let $(x_n)$ be a convergent sequence in $f^{-1}(B)$. We want to show that $\displaystyle{\lim_{n\rightarrow \infty}x_n\in f^{-1}(B)}$.
   Since $f$ is continuous, we have that $\displaystyle{f\left (\lim_{n\rightarrow \infty} x_n\right )=\lim_{n\rightarrow \infty} f(x_n)}$.
   Since $(x_n)\in f^{-1}(B)$ and $f$ is continuous we have that $f(x_n)\in B$. Is this correct? (Wondering)
   Since $B$ is closed we have that $\displaystyle{\lim_{n\rightarrow \infty}f(x_n)\in B}$.
   Therefore, we have that $\displaystyle{f\left (\lim_{n\rightarrow \infty}x_n\right )\in B}$. From that it implies that $\displaystyle{\lim_{n\rightarrow \infty}x_n\in f^{-1}(B)}$. This means that $f^{-1}(B)$ is closed.
   Is this correct? (Wondering)
   $$$$

This is ok. Thus (2) is true.

1. Since $A$ is bounded, we have that it is upper and under bounded. There are $x,y$ such that $$x\leq a \leq y , \ \ \text{ for all } a\in A$$ Since $f$ is continuous we have that $$f(x)\leq f(a) \leq f(y) , \ \ \forall a\in A$$ (no, this may not be true).
2. This means that in $A$ the function $f$ is bounded.
   Therefore, $f(A)$ is bounded.
   Is this correct? (Wondering)
   $$$$

One can show that the image of a closed interval under a continuous function $f:\mathbf R\to \mathbf R$ is again a alosed interval.
From this it will follow that (3) is true.

1. What can we do in this case? (Wondering)

As for (4), again take $f=\tan^{-1}$ and see that $f^{-1}(-\pi/2, \pi/2)=\mathbf R$. Thus (4) is false.

 

[mathmari]

One can show that the image of a closed interval under a continuous function $f:\mathbf R\to \mathbf R$ is again a alosed interval.

Isn't this the case 1., that we disaproved? (Wondering)

 

[mathmari]

Do you mean the following?

Since $A$ is bounded, there exists $b=\sup A$ and $a=\inf A$. So, $A\subseteq [a,b]$.
Since $f$ is continuous (or do we not need the continuity here ? ) we get that $f(A)\subseteq f([a,b])$.
$[a,b]$ is closed and $f$ is continuous, therefore $f([a,b])$ is closed. How could we prove that? (Wondering)
Is $f(A)$ then bounded because it is a subset of a closed interval? (Wondering)

 

[caffeinemachine]

Do you mean the following?

Since $A$ is bounded, there exists $b=\sup A$ and $a=\inf A$. So, $A\subseteq [a,b]$.
Since $f$ is continuous (or do we not need the continuity here ? ) we get that $f(A)\subseteq f([a,b])$.
$[a,b]$ is closed and $f$ is continuous, therefore $f([a,b])$ is closed. How could we prove that? (Wondering)
Is $f(A)$ then bounded because it is a subset of a closed interval? (Wondering)

In case I caused any confusion, by a closed interval I mean a subset of $\mathbf R$ of the form $[a, b]$, where $a<b$ are real numbers.

The statement I made was that if $f:\mathbf R\to \mathbf R$ is a continuous function, then $f$ maps a closed interval to a closed interval.

A one line proof of this would be to invoke the fact that any continuous image of a compact-connected topological space is again compact-connected.

Here is a more basic proof using sequences.

We want to show that $B:=f([a, b])$ is a closed interval. First we show that $B$ is bounded above. Suppose not. Then for each natural number $N$, there is $x_n\in [a, b]$ such that $f(x_n)>N$. Now by Bolzano-Weierstrass, the sequence $(x_n)$ has a convergent subsequence, say $(x_{n_k})$, which converges to say $x$. Thus $f(x_{n_k})\to f(x)$. But this means $f(x_{n_k})$ is bounded which is a contradiction. Similarly one can prove that $B$ is bounded below.

Let $x=\inf(B)$ and $y=\sup(B)$. Again, an application of Bolzano-Weierstrass shows that both $x$ and $y$ are in $B$. Now by intermediate value theorem it is clear that $B=[x, y]$.

 

[mathmari]

We want to show that $B:=f([a, b])$ is a closed interval. First we show that $B$ is bounded above. Suppose not. Then for each natural number $N$, there is $x_n\in [a, b]$ such that $f(x_n)>N$. Now by Bolzano-Weierstrass, the sequence $(x_n)$ has a convergent subsequence, say $(x_{n_k})$, which converges to say $x$. Thus $f(x_{n_k})\to f(x)$, giving $N_k\to f(x)$. But this means $N_k$ is bounded which is a contradiction. Similarly one can prove that $B$ is bounded below.

I haven't really understood the part with $N_k$. What exactly is $N_k$ ? (Wondering)

 

[caffeinemachine]

I haven't really understood the part with $N_k$. What exactly is $N_k$ ? (Wondering)

I have edited my post. (Actually, writing $N_k\to f(x)$ was not correct.)

To elaborate a bit, a subsequence of $(x_n)_{n\geq 1}$ is a sequence $(x_{n_k})_{k\geq 1}$, where $(n_k)_{k\geq 1}$ is a strictly increasing sequence of naturals.

Does that clear the confusion?

 

[mathmari]

So, we use the fact that a convergent sequence is bounded, right? (Wondering)
Having that $f(x_{n_k})$ is bounded, means that $f(x_n)$ must also be bounded? (Wondering)

Let $x=\inf(B)$ and $y=\sup(B)$. Again, an application of Bolzano-Weierstrass shows that both $x$ and $y$ are in $B$. Now by intermediate value theorem it is clear that $B=[x, y]$.

Let w be a number between x and y, then there must be at least one value c within [a, b] such that f(c) = w, right? How does it follow from that that $B=[x,y]$ ? I got stuck right now... (Wondering) In general, it holds that $B\subseteq [x,y]$, or not? Since $x,y\in B$, it follows that $B=[x,y]$, right? (Wondering)

 

[caffeinemachine]

So, we use the fact that a convergent sequence is bounded, right? (Wondering)

Yes.

Having that $f(x_{n_k})$ is bounded, means that $f(x_n)$ must also be bounded? (Wondering)

No. You can come up with easy counterexamples to this.

Let w be a number between x and y, then there must be at least one value c within [a, b] such that f(c) = w, right?

Yes.

How does it follow from that that $B=[x,y]$ ? I got stuck right now... (Wondering)

In general, it holds that $B\subseteq [x,y]$, or not? Since $x,y\in B$, it follows that $B=[x,y]$, right? (Wondering)

We defined $x=\inf(B)$ and $y=\sup(B)$. Thus we have $B\subseteq [x, y]$. The reverse inequality is by the intermediate value theorem.

 

[mathmari]

No. You can come up with easy counterexamples to this.

We want to show that $B:=f([a, b])$ is a closed interval. First we show that $B$ is bounded above. Suppose not. Then for each natural number $N$, there is $x_n\in [a, b]$ such that $f(x_n)>N$. Now by Bolzano-Weierstrass, the sequence $(x_n)$ has a convergent subsequence, say $(x_{n_k})$, which converges to say $x$. Thus $f(x_{n_k})\to f(x)$. But this means $f(x_{n_k})$ is bounded which is a contradiction.

But how do we get the contradiction?

Do we maybe have to suppose that for each sequence $x_n\in [a,b]$ there is a natural number $N$ such that $f(x_n)>N$ and not that for each natural number $N$, there is $x_n\in [a, b]$ such that $f(x_n)>N$? (Wondering)

 

[mathmari]

Consider $f(x)=\tan^{-1}(x)$. Then $f(\mathbf R)=(-\pi/2, \pi/2)$. Thus $(1)$ is not true.

How can we show that $f(\mathbf R)=(-\pi/2, \pi/2)$ ? (Wondering)

 

[mathmari]

3. $A$ bounded $\Rightarrow$ $f(A)$ bounded

Could we maybe show it also as follows?

We suppose that $f(A)$ is not bounded.

There is also a sequence $\{x_n\}_{n=1}^{\infty}$ in $f(A)$ with $\lim x_n=\infty$. Or do we not get that when $f(A)$ is not bounded? (Wondering)

Since $\{x_n\}_{n=1}^{\infty} \in f(A)$ it follows that $\{f^{-1}(x_n)\}_{n=1}^{\infty} \in A$, or not? (Wondering)

$\{f^{-1}(x_n)\}_{n=1}^{\infty}$ is therefore, bounded, right? (Wondering)

From Bolzano-Weierstrass there is a convergent subsequence, let $\{f^{-1}(x_{n_k})\}_{k=1}^{\infty}$ with $f^{-1}(x_{n_k})\rightarrow y\in \mathbb{R}$.

Since $f$ is continuous, we have that $f\left (f^{-1}(x_{n_k})\right )\rightarrow f(y)\in \mathbb{R}$.

That means that $x_{n_k}\rightarrow f(y)$.

But how could we get a contradiction? (Wondering)

 

[caffeinemachine]

Could we maybe show it also as follows?

We suppose that $f(A)$ is not bounded.

There is also a sequence $\{x_n\}_{n=1}^{\infty}$ in $f(A)$ with $\lim x_n=\infty$. Or do we not get that when $f(A)$ is not bounded? (Wondering)

It could be that there is no sequence in $f(A)$ whose limit is infinity, given that $f(A)$ is unbounded. But what is true for sure is that that is a sequence whose limit is either $+\infty$ or $-\infty$. We may WLOG assume that the limit is $+\infty$.

Since $\{x_n\}_{n=1}^{\infty} \in f(A)$ it follows that $\{f^{-1}(x_n)\}_{n=1}^{\infty} \in A$, or not? (Wondering)

Note that $f^{-1}(x_n)$ is a subset of of $\mathbf R$. So $\{f^{-1}(x_n)\}$ is not a sequence in $\mathbf R$.
So the way to go is by choose $y_n\in f^{-1}(x_n)$ for each $n$. Then you get a sequence $\{y_n\}$ in $\mathbf R$.

Now from Bolzano-Weierstrass there is a convergent subsequence $\{y_{n_k}\}_{k=1}^{\infty}$ with $y_{n_k}\rightarrow y\in \mathbb{R}$.

Since $f$ is continuous, we have that $x_n=f(y_n)\rightarrow f(y)\in \mathbb{R}$.

That means that $x_{n_k}\rightarrow f(y)$. The contradiction is that since $x_{n_k}$ is a subsequence of $\{x_n\}$, we also have $x_{n_k}\to \infty$.

 

[mathmari]

I see! Thank you very much! (Smile)