f is continuous at x 0 and f ( x 0 ) &lt; 0 then there exists a neighborhood of x 0 in which f ( x ) &lt; 0 .Is it possible to prove this statement, and if so what is the best way to go about it? Contradiction, Induction, Deduction, etc... ?

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f is continuous at       x    0   and   f  (      x    0    )  &amp;lt;  0 then there exists a neighborhood of       x    0   in which   f  (  x  )  &amp;lt;  0 .Is it possible to prove this statement, and if so what is the best way to go about it? Contradiction, Induction, Deduction, etc... ?

Ronnie Glenn · Accepted Answer

f  :  (  X  ,  d  )  →      R   continuous at       x    0    ∈  X and   f  (      x    0    )  &amp;lt;  0Choose,   ϵ  =  −            f      (              x        0              2    &amp;gt;  0Then,   ∃  B  (      x    0    ,  δ  ) such that   f  (  B  (      x    0    ,  δ  )  )  ⊆  (  f  (      x    0    )  −  ϵ  ,  f  (      x    0    )  +  ϵ  )Hence,   f  (  x  )  &amp;lt;  f  (      x    0    )  +  ϵ  =            f      (              x        0            )        2    &amp;lt;  0 for all   x  ∈  B  (      x    0    ,  δ  )If you don&#039;t know metric space, set   X  =      R   and   d  (  x  ,  y  )  =      |    x  −  y      |   and   B  (      x    0    ,  δ  )  =  (      x    0    −  δ  ,      x    0    +  δ  )

f is continuous at x 0 </msub> and f ( x 0 </msub>

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