Let f : [ 0 , 1 ] d ⟶ R d with d ≥ 2. f is continuous and let c ∈ ( 0 , 1 ). If we have that f ( 0 , . . . , 0 ) &lt;&lt; ( c , . . . , c ) and f ( 1 , . . . , 1 ) &gt;&gt; ( c , . . . , c ), is there an extension of the Intermediate Value Theorem for vector-valued functions that would help me prove that there indeed exists x ∈ ( 0 , 1 ) d such that f ( x ) = ( c , . . . , c ) ?

Question

Let   f  :  [  0  ,  1      ]          d        ⟶            R              d       with   d  ≥  2.   f is continuous and let   c  ∈  (  0  ,  1  ). If we have that   f  (  0  ,  .  .  .  ,  0  )  &amp;lt;&amp;lt;  (  c  ,  .  .  .  ,  c  ) and   f  (  1  ,  .  .  .  ,  1  )  &amp;gt;&amp;gt;  (  c  ,  .  .  .  ,  c  ), is there an extension of the Intermediate Value Theorem for vector-valued functions that would help me prove that there indeed exists   x  ∈  (  0  ,  1      )          d       such that   f  (  x  )  =  (  c  ,  .  .  .  ,  c  ) ?

Immanuel Glenn · Accepted Answer

I believe there are many things negating this, on being how would you define   a  &amp;gt;&amp;gt;  b for vectors? and with higher dimensions, we can start of &quot;below&quot; a point, and then go around it to be &quot;above&quot; it.

Let f : [ 0 , 1 ] <mrow class="MJX-TeXAtom-ORD"> d </mr

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