1
$\begingroup$

I have seen a note such as

Let $X$ and $Y$ are normed spaces. If $T:X \to Y$ is a linear isomorphism then it is linear homeomorphism but converse is not ture namely a homeomorphic mapping doesn’t have to be isometry.

Doesn’t linear isomorphism mean $T$ is linear and bijection and linear homeomorphism mean $T$ is linear, bijection, continuous and its inverse is continuous? How can we say being isomorphism requires being homeomorphism? I think its exact opposite is true.

I couldn’t be sure can someone illuminate me please?

Thank you

$\endgroup$
7
  • $\begingroup$ It's not true. Perhaps you're just not writing exactly what you saw: If $X$ and $Y$ are complete normed spaces and $T$ is a bounded linear isomorphism then $T$ is a homeomorphism. $\endgroup$ Jun 11 '18 at 17:11
  • $\begingroup$ @DavidC.Ullrich I know but if they are not complete how can we say if it is linear isomorphism then it is linear homeomorphism? $\endgroup$
    – user519955
    Jun 11 '18 at 17:27
  • $\begingroup$ I already said that that's not true. $\endgroup$ Jun 11 '18 at 17:34
  • $\begingroup$ @DavidC.Ullrich I’m so sorry. I have misunderstood.I can say if $T$ is a linear homeomorphism then it is isomorphism, can’t I? $\endgroup$
    – user519955
    Jun 11 '18 at 17:36
  • $\begingroup$ Of course not - a homeomorphism need not be linear. $\endgroup$ Jun 11 '18 at 17:38
1
$\begingroup$

To sum up @David C. Ullrich's comments with some examples:

  • If $T$ is a linear isomorphism, it doesn't have to be a linear homeomorphism. Consider these examples:

$T : c_{00} \to c_{00}$ given by $(x_n)_n = (nx_n)_n$. Here $T$ is not even bounded.

$S : c_{00} \to c_{00}$ given by $S(x_n)_n = \left(\frac1nx_n\right)_n$. Here $S$ is bounded, but its inverse $T$ is not bounded so $S$ is not a homeomorphism.

Another example: recall that $\ell^1$ and $\ell^2$ both have algebraic dimension equal to $\mathfrak c$ so there exists a linear isomorphism $\ell^1 \to \ell^2$, however there isn't a linear homeomorphism between the two spaces: namely, the dual space of $\ell^2$ is separable while the dual space of $\ell^1$ isn't. However, $\ell^1$ and $\ell^2$ are homeomorphic topological spaces.

  • If $T$ is a homeomorphism, then $T$ does not have to be a linear isomorphism. Consider $T : \mathbb{R} \to \mathbb{R}$ given by $T(x) = x^3 + x$. Here $T$ is not linear.

  • If $T$ is a linear homeomorphism, then $T$ is a linear isomorphism. This is clear since a homeomorphism is bijective by definition.

  • If $T$ is a bounded linear isomorphism between Banach spaces, then $T$ is also a linear homeomorphism. This is a consequence of the Bounded Inverse Theorem.

$\endgroup$
1
  • $\begingroup$ It is very very beneficial and clear sum thank you so much. I have given +1 $\endgroup$
    – user519955
    Jun 11 '18 at 20:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.