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Adding 2 open sets

  • Thread starter estro
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  • #1
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"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


[itex]S+T=\{s+t \| s \in S, t \in T\}[/itex]


The Attempt at a Solution


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

Will appreciate any help.
Thanks.
 
Last edited:

Answers and Replies

  • #2
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Still need help with this one...
 
  • #3
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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?
 
  • #4
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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...
 
  • #5
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3,277


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

Now, can you deduce that [itex]S(t_0+s_0,\delta)\subseteq s_0+T[/itex]??
 
  • #6
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You can find a delta such that [itex]S(t_0,\delta)\subseteq T[/itex], since T is open.

Now, can you deduce that [itex]S(t_0+s_0,\delta)\subseteq s_0+T[/itex]??
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.
 
  • #7
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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 [itex]x_0 \in S+T[/itex], so [itex]x_0=s_0+t_0[/itex] where [itex]s_0 \in S, t_0 \in T[/itex]
S is an open set so there is [itex]\delta_s>0[/itex] such that [itex]S(s_0,\delta_s) \subseteq S[/itex] so for every [itex]s \in S[/itex] that satisfies [itex]||s-s_0|| \leq \delta_s, s \in S[/itex]

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

Whats wrong?
 
Last edited:
  • #8
241
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Found my mistake.
micromass, thanks for your hint [got it at last]!
 

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