# Thread: Prove or disaprove the statements

1. 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?

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?
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?

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?

4. What can we do in this case?

2. Originally Posted by 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?

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

Originally Posted by mathmari

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?
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?

This is ok. Thus (2) is true.

Originally Posted by mathmari

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?

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.

Originally Posted by mathmari

1. What can we do in this case?
As for (4), again take $f=\tan^{-1}$ and see that $f^{-1}(-\pi/2, \pi/2)=\mathbf R$. Thus (4) is false.

Originally Posted by caffeinemachine
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?

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?
Is $f(A)$ then bounded because it is a subset of a closed interval?

5. Originally Posted by 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?
Is $f(A)$ then bounded because it is a subset of a closed interval?
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 $B=[x, y]$.

Originally Posted by caffeinemachine
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 $N_k$ ?

7. Originally Posted by mathmari
I haven't really understood the part with $N_k$. What exactly is $N_k$ ?
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?

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

Originally Posted by caffeinemachine
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...

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

9. Originally Posted by mathmari
So, we use the fact that a convergent sequence is bounded, right?
Yes.

Originally Posted by mathmari
Having that $f(x_{n_k})$ is bounded, means that $f(x_n)$ must also be bounded?
No. You can come up with easy counterexamples to this.

Originally Posted by mathmari
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.

Originally Posted by mathmari
How does it follow from that that $B=[x,y]$ ? I got stuck right now...

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

Originally Posted by caffeinemachine
No. You can come up with easy counterexamples to this.
Originally Posted by caffeinemachine
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 $f(x_n)>N$?