A<b<c and, f is bounded on [a,b]

[phydis]

Homework Statement

a<b<c and, f is bounded on [a,b] and f is bounded on [b,c] prove that f is bounded on [a,c]

The Attempt at a Solution

there exist M1≥0 s.t. for all x ε [a,b] |f(x)|≤M1
there exist M2≥0 s.t. for all x ε [b,c] |f(x)|≤M2

for x ε [a,b] and x ε [b,c]
Let M>0, and let M>M1 and M>M2
therefore
|f(x)|≤M1<M --> |f(x)|<M and |f(x)|≤M2<M --> |f(x)|<M
∴ there exist M>0 s.t. |f(x)|<M *
so f is bounded on [a,c]

is this proof correct? definition says f is bounded on [a,c] if M≥0 s.t. for all x ε [a,c] |f(x)|≤M
but what I have proven is, f is bounded on [a,c] since M>0 s.t. for all x ε [a,c] |f(x)|<M

 

[Mandelbroth]

Homework Statement

a<b<c and, f is bounded on [a,b] and f is bounded on [b,c] prove that f is bounded on [a,c]

The Attempt at a Solution

there exist M1≥0 s.t. for all x ε [a,b] |f(x)|≤M1
there exist M2≥0 s.t. for all x ε [b,c] |f(x)|≤M2

for x ε [a,b] and x ε [b,c]
Let M>0, and let M>M1 and M>M2
therefore
|f(x)|≤M1<M --> |f(x)|<M and |f(x)|≤M2<M --> |f(x)|<M
∴ there exist M>0 s.t. |f(x)|<M *
so f is bounded on [a,c]

is this proof correct? definition says f is bounded on [a,c] if M≥0 s.t. for all x ε [a,c] |f(x)|≤M
but what I have proven is, f is bounded on [a,c] since M>0 s.t. for all x ε [a,c] |f(x)|<M

Uh...if $x\in[a,b]$ and $x\in[b,c]$, then $x=b$. I think you meant $x\in[a,c]$.

 

[phydis]

Uh...if $x\in[a,b]$ and $x\in[b,c]$, then $x=b$. I think you meant $x\in[a,c]$.

Nope, I meant some $x\inℝ$ lies on [a,b] and some $x\inℝ$ lies on [b,c]

 

[Mandelbroth]

Nope, I meant some $x\inℝ$ lies on [a,b] and some $x\inℝ$ lies on [b,c]

That's not a good way to put it. It's confusing. Try using $x_1$ and $x_2$.

 

[HallsofIvy]

What you should say is "if x\in [a, c] then either x\in [a, b] or x\in [b, c]". (You shouldn't use "x" to mean two different numbers.)

Also where you say "Let M>0, and let M>M1 and M>M2" it looks as if you were "letting" M be three different numbers. Better would be "Let M> max(M1, M2)". Of course, since M1 and M2 are both positive, it follows that M> 0.