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{z ∈ $\mathbb{C}$ | Re(z) < −1 or Re(z) > 4}

Show whether it is open, closed or neither and if it is connected

Own work:

The set can be expressed as:

{z ∈ $\mathbb{C}$ | Re(z) < −1} ∪ {z ∈ $\mathbb{C}$ | Re(z) > 4}

These two sets are seen to be non-empty, open and disjoint hence not connected. Alternatively, any path from a Re(z)>4 and Re(z)<-1 must pass through a point whose real part is 0, contradicting the Intermediate Value theorem.

Is this correct? And how do I show the two sets expressed are open if they are so?

1 Answers 1

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Yes, that's correct, to show that these sets are open let $Re(z)>4$ take $\varepsilon=Re(z)-4$, then $Re(z_0)>4$ for any $z_0\in B(z,\varepsilon)$. The other one is similar and easy if you understand this case.

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    To show the other one is open is the proof, Re(z)<-1 take ε=Re(z)+1 , then Re(z)<-1 for any z∈B(z,ε). is this correct? I think i followed a similar approach to your proof – 2017-02-24
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    I think $\varepsilon=-1-Re(z)$ would be the appropriate value for $\varepsilon$. Remember that $\varepsilon$ must be positive, the value you chose is not. – 2017-02-24