Let $X$ be a path-connected topological space and assume for a contradiction that $X=A \cup B$, where $A \cap B = \emptyset$ and $A,B \neq \emptyset$ (i.e. $X$ isn't connected). Choose $a \in A$, $b \in B$ Let $f: [0,1] \rightarrow X$ be a path in $X$ such that $f(0)=a$, $f(1)=b$. Then by continuity of $f$ there is a decomposition $$[0,1]=f^{-1}(A) \cup f^{-1}(B)$$ and this decomposition implies that $[0,1]$ is not connected, which is not true. Hence we conclude that $X$ cannot be path-connected, if it isn't connected.

I am studying Real analysis and don't really understand everything in topology. I will write all the troubles that I am having.

1) When we say that inverse image of an set, Is it necessary that all points in the set has to have inverse ? (In this case we took inverse of A where all points in A did not have its inverse.

2) So If 1) is true , then I am having trouble in understanding this theorem that if f is continuous function then inverse image of any open set is open set.

If f is a continuous function from [0,1] to [1,2] , Now If I take inverse image of (0,3) it should be an open set since it is inverse image of an open set under continuous function .But inverse image in this case is [0,1] which is closed .

So what is that I am missing ?


1) The inverse image of a set $A\subset Y$ under a map $f:X\to Y$ is the set of all elements of $X$ that $f$ sends to $A$. So:

$$f^{-1}(A)=\{x\in X: f(x)\in A\}$$

2) This is not a theorem -- it is the definition of a continuous function. A function $f:X\to Y$ is continuous (by definition) if the inverse image of every open set in $Y$ under $f$ is open in $X$.

3) In your final example, you have take care with the topology of $(0,3)$. It is not open in the range of $f$.

  • $\begingroup$ It might be treated as definition in topology but in real analysis delta -epsilon definition is used. $\endgroup$ – Shubham Ugare Sep 8 '15 at 17:31
  • $\begingroup$ That's why it is best to study topology first :) Anyway, epsilon-delta is used when there are metrics around, and metrics of course induce topologies, and then it is very easy to see that the two definitions are equivalent. $\endgroup$ – uniquesolution Sep 8 '15 at 17:34

The notation $f^{-1}(A)$ means the values $y\in [0,1]$ so that $f(y)\in A$. It doesn't imply that every point $c\in A$ has an inverse... i.e. some $y\in [0,1]$ so that $f(y)=c$.

As to the second question, the interval $[0,1]$ is an open set if that is the whole space. I.e. $[0,1]$ is an open set in $[0,1]$, but $[0,1]$ is not an open set in $\mathbb{R}$.

  • $\begingroup$ Okay .I understood the mistake . $\endgroup$ – Shubham Ugare Sep 8 '15 at 17:45

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