What is a nice example for non-deformable sets in a topological space? Let $M$ be a separable topological space. A set $X \subset M$ is said to be deformable in $M$ into $Y \subset M$ if there is a continuous mapping $f_0$ from $X$ to $Y$ which is homotopic in $M$ to identity map of $X$ in $M$. 
I am not able to think of any non-examples of deformable sets. Is $(0,1)$ deformable into 2 element set {$0,1$} ?
What are some nice real world examples of non-deformable sets and deformable sets ? Are there any non-deformable subsets of $\mathbb{R}$ ?
 A: Any subset $X \subset \mathbb R$ is deformable into any nonempty subset $Y \subset \mathbb R$, because you can take any $y \in Y$ and then define the homotopy $h : X \times [0,1] \to \mathbb R$ by
$$h(x,t) = (1-t)x + t y
$$
A similar idea shows that for any contractible space $M$, any subset $X \subset M$ is deformable into any nonempty subset $Y \subset M$.
On the other hand, consider $M = \mathbb{R} - \{0\} = (-\infty,0) \cup (0,\infty)$, let $X \subset M$, and suppose that $X$ contains a point of $(-\infty,0)$ and a point of $(0,\infty)$. In this situation $X$ is not deformable into $\{1\}$, and more generally $X$ is deformable into $Y \subset \mathbb R$ if and only if $Y$ contains a point of $(-\infty,0)$ and a point of $(0,\infty)$.
What I'm doing with these two examples is I am employing the homotopy invariants of algebraic topology to give me examples and counterexamples. So far, I've only used the set of path components, also known as $\pi_0$.
Here is a $\pi_1$ example. If $M = \mathbb R^2 - \{0\}$ and $X = S^1 \subset M$, then $X$ is not deformable into any one point subset $Y \subset M$, because if it were then the inclusion induced isomorphism $\mathbb Z \approx \pi_1(S^1) \approx \pi_1(M) \approx \mathbb Z$ would factor through $\pi_1(Y)=0$, which is clearly impossible.
