Proving S+T is Open Set: Step-by-Step

[estro]

"Adding" 2 open sets

Homework Statement

I'm trying to prove that If both S and T are open sets then S+T is open set as well.

Homework Equations

S+T=\{s+t \| s \in S, t \in T\}

The Attempt at a Solution

S+T is open if every point x_0 \in S+T is inner point.
Let x_0 be a point in S+T, so there is s_0 in S and t_0 in T so that x_0=s_0+t_0.
S is open so for every ||s-s_0|| < δ_1 s in S.
T is open so for every ||t-t_0|| < δ_2 t in T.

Let x be point in S(x_0, _delta_), I will write x=s+t. [both s and t are some vectors in R^n]
s+t in S(x_0, _delta_)={s+t | ||s+t-s_0-t_0|| < _delta_} and here I stuck, if I could conclude from ||s+t-s_0-t_0|| < _delta_ that ||s-s_0|| < δ_1 and ||t-t_0|| < δ_2 the proof will be over, however I just can't find the algebraic manipulation.

Will appreciate any help.
Thanks.

 

[estro]

Still need help with this one...

 

[micromass]

Please don't bump after only 12 hours. Wait at least 24 hours.

Anyway. Take s in S fixed. Can you prove that s+T is open?

 

[estro]

Sorry for being impatient.

When I try to prove that s_0+T is open I get into the same trouble:
s_0+T is open set if there is δ such that S(x_0=s_0+t_0, δ) in s_0+T, so let x_0 be point in S(x_0=s_0+t_0, δ) but from the definition: S(x=s_0+t_0, δ)={x | ||s_0+t_0-x||<δ}, in short I get into the same kind of trouble...

 

[micromass]

You can find a delta such that S(t_0,\delta)\subseteq T, since T is open.

Now, can you deduce that S(t_0+s_0,\delta)\subseteq s_0+T??

 

[estro]

You can find a delta such that S(t_0,\delta)\subseteq T, since T is open.

Now, can you deduce that S(t_0+s_0,\delta)\subseteq s_0+T??

I think I was able to prove that s_0+t is open:

Let x_0 be in s_0+T, so there is t_0 and s_0 such that s_0 in S and t in T.
S is open so there is δ_1 so that S(s_0, δ_1) in S, so if ||s-s_0||< δ_1 then s in S.
Let x be in S(s_0+t_0, δ_1)={x=s_0+t | ||s_0+t-s_0-t_0|| < δ_1, s_0 in S}={x=s_0+t | ||t-t_0|| < δ_1, s_0 in S}, now because ||t-t_-0||< δ_1 t is in T so S(s_0+t_0, δ_1) in s_0+T.

Is this ok?
Now trying to prove the more that S(t_0+s_0,_some_delta_) in S+T.

 

[estro]

I think the above proof is wrong because with similar technique I can prove that S+T is open set even if only S {or T} are open:

Let x_0 \in S+T, so x_0=s_0+t_0 where s_0 \in S, t_0 \in T
S is an open set so there is \delta_s&gt;0 such that S(s_0,\delta_s) \subseteq S so for every s \in S that satisfies ||s-s_0|| \leq \delta_s, s \in S

Let x \in R^n, we can write it as x=t_0+s where t_0 \in T, s \in R^n then if x=t_0+s \in S(s_0+t_0, \delta_s)=\{t_0 + s | \|t_0+s-t_0-s_0\|&lt; \delta_s, t_0 \in T\}=\{t_0+s | \|s-s_0\|&lt; \delta_s, t_0 \in T\} \subseteq S+T because t_0 \in T\ and\ s \in S

Whats wrong?

 

[estro]

Found my mistake.
micromass, thanks for your hint [got it at last]!