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I am trying to understand the nuances among different possibly correct definitions over some core concept in calculus/ real analysis to deepen my understanding. I attempted some true/false questions. Let me know if I am correct. NOTE: let me know if I am wrong, I will try to justify and pinpoint my confusion! Thanks a ton for people helped!
1. True, it's stardard defintion I learn so it's obvious

2.F. Because there is no 0<|x|

3.True. I feel like its true..

4.False Because 5 is true

5.T. Because the preimage of an open set is open

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this discussion of ( math.drexel.edu/~sar323/exposition/opensets.pdf ) of open and closed sets and their relation to continuity might help you cement your understanding. –  toypajme Dec 19 '12 at 5:19

1 Answer 1

up vote 1 down vote accepted

The definition of continuity at $0$ is $$\forall \epsilon>0\exists \delta>0:\left|x\right|<\delta\Rightarrow \left|f(x)-f(0)\right|<\epsilon$$ There is no $0<\left|x\right|$!!! (this is added when talking about limits).

Therefore, 1 and 2 are both true.

3 is true if and only if $f(0)=0$!! (generally $a_n\to 0\Rightarrow f(a_n)\to f(0)$ whenever $f$ is continuous at $0$)

4 is not generally true but not because of your reasoning

5 is true

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Oh, yes, so the first one is false, the second is true. Right? What about the others? And my reasoning? Thanks a lot! –  user48601 Dec 17 '12 at 17:52
    
No! The first one is true as well because $2\Rightarrow 1$. –  Nameless Dec 17 '12 at 17:53
    
Okay, thank you! I'd better do another round. Gush. Reals hard. –  user48601 Dec 17 '12 at 17:57
1  
@Y.F.Stewart Before you attempt anything, make sure you know exactly the definitions... –  Nameless Dec 17 '12 at 17:59
    
Yes, that's why I am doing this now! Thank you, Nameless lifesaver! –  user48601 Dec 17 '12 at 18:12

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