Vtyjkyuk

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 $mathbbR$ ?

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 $M = mathbbR - 0 = (-infty,0) cup (0,infty)$, 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.